InterviewSolution
Saved Bookmarks
InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 2801. |
Give two examples of reversible processes. Discuss their reversibility? |
|
Answer» (i) Melting as well as vaporistion are reversible processes. On reversing the conditions under which they occur, the vapour condence into a liquid and liquid solidifies. (ii) All isothermal and adiabatic processes, in which no extraneous loss of heat occurs, can be retraced by reversing the boundry conditions and hence they are revesible processes. |
|
| 2802. |
Two moles of an ideal gas expanded isothermally and reversibly from `1L` to `10L` at `300K`. What is the enthalpy change?A. 4.98kJB. 11.45 KJC. `-11.45kJ`D. 0 kJ |
|
Answer» Correct Answer - d |
|
| 2803. |
Two moles of an ideal gas expanded isothermally and reversibly from `1L` to `10L` at `300K`. What is the enthalpy change?A. `4.98 kJ`B. `11.47 kJ`C. `-11.47 kJ`D. `0 kJ` |
| Answer» Temperature remains constant, so enthalpy change for an isothermal process is equal to zero. | |
| 2804. |
`1 g` mole of an ideal gas at STP is subjected to a reversible adiabatic expansion to double its volume. Find the change in internal energy `( gamma = 1.4)` |
| Answer» Correct Answer - `-1373.2J` | |
| 2805. |
2 mole of zinc is dissolved in HCl at `25^(@)` C. The work done in open vessel is : |
| Answer» Correct Answer - `-4.955 kJ` | |
| 2806. |
Classify the following processes as exothermic or endothermic: (A) Burning of a match stick (B) Melting of ice (C ) Molten metal solidifies (D) Reaction between `Na` and `H_(2)O` (E ) Rubbing alcohol evaporates. |
| Answer» Correct Answer - Exothermic `A,C,D` ; endothermic ; `B,E` | |
| 2807. |
Which of the following statement (s) is correct? Statement-I: The entropy of isolated system with P-V work only, is always maximized at equillibrium. Statement-2: It is possible for the entropy of closed system to decrease substantialy in an irreversible process. Statemet-3: Entropy can be crearted but not destroyed. Statement-4 `DeltaS_(system)` is zero for reversible process in an isolated system.A. Statement `(I, ii,iii)`B. Statement `ii, iv`C. Statement `I ,ii,iv`D. All of these |
| Answer» Correct Answer - D | |
| 2808. |
Select the correct statenent (s) : `S_(1) : AICI_(3)` when dissolve in `H_(2)O` its entropy increases althrough it is a spontaneous process . `S_(2):` When `H_(2)` gas adsrobed at the surface of `Pd` , some amount of heat is released. `S_(3):` Entropy of `D_(2)` gas is greater than `H_(2)` gas.A. `S_(1), S_(2) & S_(3)`B. `S_(1) "&" S_(3)`C. `S_(2) "&" S_(3)`D. `S_(1) "&" S_(2)` |
| Answer» Correct Answer - C | |
| 2809. |
There is no change in internal energy for an ideal gas at constant temperature. Internal energy of an ideal gas is a function of temperature only.A. If both the statements are TRUE and STATEMENT-2 is the correct explanation of STATEMENT-2B. If both the statements are TRUE but STATEMENT-2 is NOT the correct explanation of STATEMENT-2C. If STATEMENT-1 is TRUE and STATEMENT-2 is FALSED. If STATEMENT-1 is FALSE and STATEMENT-2 is TRUE |
| Answer» Correct Answer - A | |
| 2810. |
Which of the reaction defines molar `DeltaH_(f)^(@)` ?A. `N_(2)(g)+3H_(2)(g)to2NH_(3)(g),H_(25^(@)C)^(@)=-92.2 kJ`B. `(1)/(2)Br_(2)(g)=(1)/(2)H_(2)(g)toHBr(g)`C. `N_(2)(g)+2H_(2)(g)+(3)/(2)O_(2)(g)toNH_(4)NO_(3)(s)`D. `I_(2)(s)=H_(2)(g)to2HI(g)` |
|
Answer» Correct Answer - C |
|
| 2811. |
Heat and work are 'definite quantities'. Heat and work are not properties of a system. Their values depend on the path of the process and vary accordingly.A. If both the statements are TRUE and STATEMENT-2 is the correct explanation of STATEMENT-1B. If both the statements are TRUE but STATEMENT-2 is NOT the correct explanation of STATEMENT-1C. If STATEMENT-1 is TRUE and STATEMENT-2 is FALSED. If STATEMENT-1 is FALSE and STATEMENT-2 is TRUE |
| Answer» Correct Answer - D | |
| 2812. |
In the reaction, `CO_(2)(g)=H_(2)(g)toCO(g)=H_(2)O(g)," "DeltaH=2.8 kJ` `DeltaH` represents :A. Heat of reactionB. Heat of combustionC. Heat of formationD. Heat of solution |
|
Answer» Correct Answer - A |
|
| 2813. |
Which of the following reactions are endothermic reactions ?A. Combustion of methaneB. Decomposition of waterC. Dehydrogenation of ethaneto etheneD. Conversion of graphite to diamond |
|
Answer» Correct Answer - b,c,d Combustion of methane is exothermic. All other given reactions are endothermic. |
|
| 2814. |
A natural gas may be assumed to be a mixture fo methane and ethane only. On complete combustion of `10L` of gas at `STP` the heat evolved was `474.6 kJ`. Assuming `Delta_(comb) H^(Theta) CH_(4)(g) =- 894 kJ mol^(-1)` and composition of the mixture by volume. |
|
Answer» `xL rarr CH_(4), "mole of" CH_(4) = x //22.4` `(10-x) L rarr C_(2)H_(6), "mole of" C_(2)H_(6) = (10-x)//22.4` Heat evolved `= (x)/(22.4) xx 894 +((10-x))/(22.4) xx 1500` `474.6 = (x)/(22.4) xx 894 +((10-x))/(22.4) xx 1500` `x = 0.745, % CH_(4) = 74.5%` |
|
| 2815. |
Select the correct statement from the following:A. In exothermic reaction, the value of equilibrium constant increases with rise of temperature.B. In endothermic reaction, the value of equilibrium constant decreases with rise in temperature.C. In exothermic reaction, the value of equilibrium constant decreases with rise of temperature.D. In endothermic reaction, the value of equilibrium constant remains constant with rise of temperature. |
| Answer» During exothermic reaction, heat evolved on reaction proceeding forwed direction. When the temperature increases of the system the reaction will proceed in backward direction. | |
| 2816. |
A system goes from P to Q by two different pats in the P-V diagram as shown in Fig. Heat given to the system in path 1 is 1000 j. the work done by the system along path 1 is more than path 2 by 100 J. What is the heat exchanged by the system in path 2? |
|
Answer» For path 2, ∆Q = ∆U + ∆W For path 1, 1000 = ∆U + (∆W + 100) or, ∆U = ∆W = 900 or, ∆Q = 900 J |
|
| 2817. |
A thermodynamic system is taken from an original state D to an intermediate state E by the linear process shown in (figure) Its volume is then reduced to the original value from E to D via F by an isobaric process. Calculate the total work done by the gas from D to E to F to D. |
|
Answer» `"Total work done by the gas from" D " to " E " to " F = "Area of" DeltaDEF` `"Area of" DeltaDEF=(1)/(2)DExxEF` Where, DF = Change in pressure `=600N//m^(2)-300N//m^(2)` `=300N//m^(2)` FE = Change in volume `=5.0m^(3)-2.0m^(3)` `=3.0m^(3)` `"Area of" DeltaDEF=(1)/(2)xx300xx3=450J` Therefore, the total work done by the gas from D to E to F is 450 J. |
|
| 2818. |
The heat absorbed by the system in going through the cyclic process as shown in figure is A. 30.4 JB. 31.4 JC. 32.4 JD. 33.4 J |
|
Answer» Correct Answer - B In a cyclic process heat absorbed by the system is equal to the work done by the system `therefore` Hence absorbed =work done =area of the given circle `=pir^(2)=pixx((Delta,P)/(2))((DeltaV)/(2))` =`3.14xx((200)/(2))xx10^(3)xx((200)/(2))xx10^(-6)` `=3.14xx100xx10^(3)xx10xx10^(-6)=31.4 J` |
|
| 2819. |
Titanium metal is extensively used in aerospace industry because the metal imparts strength to structures but does not unduly add to their masses. The metal is produced by the reduction of `TiCl_(4(l))` which in turn is produced from mineral rutile `[TiO_(2(s))]`. can the following reaction for production of `TiCl_(4(l))` be carried out at `25^(@)C`? `TiO_(2(s))+2Cl_(2(g))toTiCl_(4(l))+O_(2(g))` Given that `H_(f)^(@)` for `TiO_(2(s)),TiCl_(4(l)),Cl_(2(g))` and `O_(2(g))` are `-944.7, -804.2,0.0,0.0 kJ mol^(-1)`. also `S^(@)` for `TiO_(2(g)),TiCl_(4(l)),Cl_(2(g))` and `O_(2(g))` are `50.3,252.3,233.0,205.1 J mol^(-1) K^(-1)` respectively. |
| Answer» Correct Answer - `158.06 kJ` | |
| 2820. |
An aeroplane weighing `63,000 kg` flies up from sea level to a height of `8000` meter. Its engine run with pure normal octane `(C_(8)H_(18))` has a `30%` efficiency. Calculate the fuel cost of the flight if octane sells at `Rs. 3` per litre. Given density of octane `=0.705 g mL^(-1)`, heat of combustion of octane `=1300 kcal mol^(-1) (g=981 m//s^(2))` |
| Answer» Correct Answer - `1472.4 Rs` | |
| 2821. |
In `P-V` diagram shown below, A. `AB` represents adiabatic process.B. `AB` represents isothermal process.C. `AB` represents isobaric process.D. `AB` represents isochoric process. |
| Answer» In `AB` process, volume does not change hence it is isochoric process. | |
| 2822. |
Given that `Cu_(4)(g) +360 kJ rarr C(g) +4H(g)` `C_(2)H_(6)(g) +620 kJ rarr 2C(g) +6H(g)` The value of `C-C` bond enegry isA. `260 kJ mol^(-1)`B. `180 kJ mol^(-1)`C. `130 kJ mol^(-1)`D. `80 kJ mol^(-1)` |
|
Answer» `CH_(4)rarr C(g) +4H(g),DeltaH = 360 kJ` `C_(2)H_(6)rarr 2C(g)+6H(g), DeltaH = 620 kJ` `:. DeltaH_(C-C) = (DeltaH_(C_(2)H_(6))-6DeltaH_(H)) = 80 kcal mol^(-1)` |
|
| 2823. |
Which of the following statements is/are correct? A. `A` represents isochroic processB. `B` represents adiabatic processC. `C` represents isothermal processD. `D` represents isobaric process |
| Answer» Correct Answer - a,b,c,d | |
| 2824. |
Consider the modes of transformations of a gas form state `A` to state `B` as shown in the given `P-V` diagram. Which one of the following is true? A. `DeltaH = q` along `A rarrC`.B. `Delta S` is same along both `A rarrB` and `A rarr C rarrB`C. `w` is same along both `A rarrB` and `A rarr C rarrB`D. `w gt 0` along both `A rarr B` and `A rarrC` |
| Answer» Work `(w)` is a state function. It depends upon the initial and final state of system. | |
| 2825. |
If the heat fo dissolution of anhydrous `CuSO_(4)` and `CuSO_(4).5H_(2)O` is `-15.89 kcal` and `2.80 kcal`, respectively, then the heat of hydration fo `CuSO_(4)` to form `CuSO_(4).5H_(2)O` isA. `-13.09 kcal`B. `-18.69 kcal`C. `+13.09 kcal`D. `+18.69kcal` |
|
Answer» `CuSO_(4)(s)+aq rarr CuSO_(4).5H_(2)O,DeltaH^(Theta) =- 15.89 kcal` `CuSO_(4).5H_(2)O +aq rarr CuSO_(4).5H_(2)O,DeltaH^(Theta) =- 2.80 kcal` Heat of hydration `=- 15.89 - 2.80 =- 18.69 kcal` |
|
| 2826. |
Consider a reaction :`A(g)+B(g)hArrC(g)+D(g)` A(g), B(g) and C(g) are taken in a container at 1 bar partial pressure each and adequate amount of liquid D is added. From the data give below calculate four digit number abcd. Givin : `DeltaG_(f)^(@)A(g)=30kJ//"mole",` `DeltaG_(f)^(@)B(g)=20kJ//"mole",` `DeltaG_(f)^(@)C(g)=50kJ//"mole",` `DeltaG_(f)^(@)D(g)=100kJ//"mole",` Vapour pressure of G (l) at `300K=(1)/(6)"bar"` (All data at 300K) where Equilibrium constant of reaction (i) Twice the partial pressure of B at equilibrium Twice the partial pressure of B at equilibrium Twice the partial pressure of C at equilibrium |
|
Answer» Correct Answer - 1113 |
|
| 2827. |
Calculate enthalpy change (in Kj) when 2 mole of liquid acetic acid undergoes dissociation into `CH_(4)(g)` and `CO_(2)(g)` from the following date : `DeltaH_(vap)[CH_(3)COOH](l)=50kJ//"mole"` Resonance energy of `DeltaH_(vap)[CH_(3)COOH](g)=-50kJ//"mole"` Resonance energy `(kJ//"mole"):C-H=400,C-O=350,o=o=500` `C-C=350,O-H=450, C=O=800,H-H=400` |
|
Answer» Correct Answer - 50 |
|
| 2828. |
A refrigerator is to maintain eatables kept inside at `9^(@)C`, if room temperature is `36^(@)C`. Calculate the cofficient of performance. |
|
Answer» Temperature inside the refrigerator, `T_(1)=9^(@)C=282K` Room temperature, `T_(2) = 36^(@)C = 309 K` Coefficient of performance`=(T_(1))/(T_(2)-T_(1))` `=(282)/(309-282)` `=10.44` Therefore, the coefficient of performance of the given refrigerator is 10.44. |
|
| 2829. |
A refrigerator is to maintain eatables kept inside at `7^(@)C`. The coefficient opf performance of refrigerator if room temperature is `387^(@)C` isA. 15.5B. 16.3C. 20.1D. 9.03 |
|
Answer» Correct Answer - D Here , `T_(1)=38^(@)C=38 + 273 = 311 K` `T_(2)=7^(@)C = 7 + 273 =280 K` `therefore` coeffiecient of performance of the refrigeratior, `=(T_(2))/(t_(1)-T_(2))=(280)/(311-280)`=9.03 |
|
| 2830. |
A thermodynamic system is taken form an original state to an intermediate state by the linear process shown in (Fig.) Its volume is then reduced to the original value form E to F by an isobaric process. Calculate the total work done by the gas from D to E to F. |
|
Answer» As it is clear from the above figure, Change in pressure, dp = EF = 5.0 - 2.0 = 3.0 atm = 3.0 x 105 Nm-2 Change in volume, dV = DF = 600 -300 = 300 c.c = 300 x 10-6 m3 Work done by the gas from D to E to F = Area of ΔDEF w 1/2 = DF x EF = (300 X 10-6) x (3.0 x 105) = 45 J. |
|
| 2831. |
Calculate `DeltaG_(reaction) ("kJ"//"mol")` for the given reaction at 300 K `A_(2)(g)+B_(2)(g)hArr2Ab(g)` and at particle pressure of `10^(-2)`bar and `10^(-4)` Given : `Delta H_(f)^(@) AB =180 kJ//mol," "DeltaH_(f)^(@) A_(2)=60 kJ//mol` `Delta H_(f)^(@) B_(2) = 29.5 kJ//mol," "DeltaS_(f)^(@) AB=210 J//K-mol` `Delta S_(f)^(@) A_(2) = 190 kJ//mol," "DeltaS_(f)^(@) B_(2)=205 J//K-mol` Use :` 2.303 Rxx300=5750 "J"//"mole" ` |
|
Answer» Correct Answer - 234 |
|
| 2832. |
The heat evolved on combustion of 1 gm of starch, `(C_(6)H_(10)O_(5))x`, into `CO_(2)(g) " and " H_(2)O(l)` at constant pressure, is 4.00 kcal. Standard Ethalpy of formation of `CO_(2)(g)" and " H_(2)O(l)` are `-94.00 " and " -65.40 "kcal"//"mol"`. The magnitude of standard enthalpy of formation of starch (in `al//gm`)is : |
|
Answer» Correct Answer - 1500 |
|
| 2833. |
An imaginary engine, is capable of expanding the gas upto `10^(13)` times. If the engine expands the gas upto the maximum possible extent isothermally at 300 K, then work done by gas is x, whereas if the engine expands upto the maximum possible extent adiabatically, at an initial temperature of 300 K, the temperature falls to -160.6 Kand the work done by gas is y. If the gas is He and all process are reversible in nature, calculate by how many times x is greater than y. |
|
Answer» Correct Answer - 13 |
|
| 2834. |
12.5 millinmole of `NH_(4)NO_(3)` dissolved in enough water to make 25.0 mL of solution. The initial temperature is `25.8^(@)C` and temperature after solid dissolves is `21.8^(@)C`. Calculate the enthalpy of solution for the `NH_(4)NO_(3)(s) "in kcal"//"mole"`. [Given : Density of the solution `=1 "gm"//"ml"` and heat capacity of solution `(1cal)/(gm-K)`] |
|
Answer» Correct Answer - 8 |
|
| 2835. |
Once mole of gas is subjected to a process causing a change in astate from(1.25 atm, 300 K) to a final state of (1 atm, 600 K). Calculate the enthalpy change from the following information [in atm-litre] Information 1: The process involvea 100 atm-litre of heat given to system of 20 atm-liter of work is done by the system. Information :2 Molar mass of the gas is 49.26. Information 3 : Density of gas as 1.25 atm and 300 K is 2 gm/liter. Information 4: Density of gas at 1 atm and 600 K is l gm/litre. [Given : R =0.0821 atm-litre/mol `K=(18.47)/225` atm-litre/mole K] (Round off your answer to nearest integer). |
|
Answer» Correct Answer - 98 |
|
| 2836. |
Calculate `Delta U` reaction for the hydrogenation of acetalence at constant volume and at `77^(@)C`. Given that `-DeltaH_(f)(H_(2)O)= -678` kcal mole , `Delta H_("comb")(C_(2)H_(2))= -310.1 kcal//ms^(2)` |
|
Answer» Correct Answer - 40 Required equation is : `C_(2)H_(2(g)) + H_(2(g)) rarr C_(2)H_(4(g)) " "DeltaH-DeltaU-=?` `H_(2(g))+ (1)/(2)O_(2(g)) rarr H_(2)O(l) " " DeltaH- = -67.8 " ""…"`(1) `C_(2)H_(2(g)) + (5)/(2)O_(2(g)) rarr 2CO_(2) + 2H_(2)O(l)" " DeltaH-=-310.1" ""..."`(2) `C_(2)H_(4) + 3O_(2(g)) rarr 2CO_(2(g)) + 2H_(2)O(l)" " DeltaH-=-337.2 " ""..."(3)` `(1) + (2) -(3)` `implies C_(2)H_(2(g)) rarr C_(2)H_(4(g))` `DeltaH_(rxxn) = -67.8 - 310.1 + 337.2 = -40.7 KCal` `DeltaH = DeltaU + Deltan(g)RT` `-40.7= DeltaU +(-1) xx 2 10^(-3) xx 350` `DeltaU= -40.7+.7` `DeltaU=-40"KCal"//"mole"` |
|
| 2837. |
Calculate resonance energy for `(1)/(6)` mole of naphtalene if its heat of hydrogenation is 91 kcal and heat of hydrogenation of id 29 kcal. |
|
Answer» Correct Answer - 9 |
|
| 2838. |
An ideal gas undergoes a process such that `P prop(1)/(T)`. If molar heat capacity for this process is `C =33.24" J"//"mole-K"`, then calculate A. Where `A = 2gamma " and " gamma` is adiabatic index of gas. `(R= 8.31 "J"//"mole-K")` |
|
Answer» Correct Answer - 3 |
|
| 2839. |
For the reaction `Ag_(2)O(s) rarr 2Ag(s) + (1)/(2) O_(2)(g) , DeltaH ` is `30.56kJ mol^(-1)` and `Delta S ` is `66JK^(-1) mol^(-1)` at one atmosphere pressure. Calculate the temperature at which `Delta G` for it will be zero. What will be the direction of the reaction at this temperature and at temperature above or below this temperature and why ? |
|
Answer» Correct Answer - `DeltaG = 0 ` at 463K . At this temp, the reaction will be in equilibrium . Above this temperature , the reaction will be spontaneous in the forward direction . Below this temp, the reaction will be non-spontaneous or it wll proceed in the bakcward direction. When `DeltaG = 0, T = (DeltaH )/(DeltaS) = ( 30560J mol^(-1))/( 66JK^(-1)mol^(-1))= 463.0K` `DeltaG = Delta H - T DeltaS `. Above 463K `, DeltaG = -ve` ( because `DeltaH ` and `Delta S` both are `+ve`) . Below 463K, `DeltaG = + ve`. |
|
| 2840. |
The temperature at which the reaction `Ag_(2)O(s) rarr 2Ag(s) +1/2 O_(2) (g)` at 1 atmospheric pressure will be in equilibrium is ____ K. The value of `DeltaH` and `DeltaS` for the reaction are 30.58 kJ and `66.11 JK^(-1)` respectively and these value do not change much with temperature.A. 462.6 KB. 486.4 KC. 364.5 KD. 521.2 K |
|
Answer» Correct Answer - A `DeltaG=DeltaH-T Delta S` |
|
| 2841. |
Gibbs energy of Gibbs function, `G`, is defined asA. `G = U + TS`B. `G = H - TS`C. `G = H + TS`D. `G = U - TS` |
|
Answer» Correct Answer - B Gibbs energy is the part of total energy which is available for doing useful work. `H` stands for total energy of the system while `TS` stands for energy for nonuseful work (or the energy which is not available to do useful work). |
|
| 2842. |
If `Delta H_("vap")` of pure water at `100^(@)C` is `40.627 kJ mol^(-1)`. The value of `Delta S_("vap")` isA. `10.8 91 kJ mol^(-1)`B. `108.91 Jk^(-1) mol^(-1)`C. `606.27 JK^(-1) mol^(-1)`D. `808.27 JK^(-1) mol^(-1)` |
|
Answer» Correct Answer - B `DeltaS=(DeltaH)/(T)` |
|
| 2843. |
Which of the following permits the calculation of absolute values of entropy of a pure substance from thermal data alone?A. Second law of thermodynamicsB. Third law of thermodynamicsC. First law of thermodynamicsD. Zeroth of thermodynamics |
|
Answer» Correct Answer - B For a pure substance, this can be done by summing `q_(rev)` increments from `0 K` to `298 K`. Along with this information, we use the fact that the entropy of any pure, perfect crystalline substance approaches zero as the temperature approaches absolute zero. |
|
| 2844. |
The enthalpy and entropy change for the reaction, `Br_(2)(l)+Cl_(2)(g)rarr2BrCl(g)` are `30KJmol^(-1)` and `105JK^(-1)mol^(-1)` respectively. The temperature at which the raction will be in equilibrium is:A. 300 KB. 285.7 KC. 273 KD. 450 K |
|
Answer» Correct Answer - B `DeltaG=DeltaH-T Delta S` |
|
| 2845. |
The enthalpy and entropy change for the reaction, `Br_(2)(l)+Cl_(2)(g)rarr2BrCl(g)` are `30KJmol^(-1)` and `105JK^(-1)mol^(-1)` respectively. The temperature at which the raction will be in equilibrium is:A. `450 K`B. `300 K`C. `285.7 K`D. `273 K` |
|
Answer» Correct Answer - C According to thermodynamics, `Delta G = Delta H - T Delta S` At equilibrium (by definition), `Delta G = 0`. Thus, ` 0 = Delta H - T Delta S` or `T = (Delta H)/(Delta S)` `= (30 xx 10^(3) J mol^(-1))/(105 J K^(-1) mol^(-1))` `285.7 K` |
|
| 2846. |
How much heat (in joule) is required to raise the temperature of `205 g` of water from `21.2^(@)C` to `91.4^(@)C`. Specific heate of water is `4.18 J g^(-1) .^(@)C^(-1)` Strategy : The specific heat of a substance is the amount of heat required to raise the temperature of `1g` of the substance by `1^(@C`. Thus Specific heat `= (("Amount of heat in joules"))/(("Mass of substance in grams")("Temperature change in" .^(@)C))` Rearrangement gives. Amount of heat `(q) =` Mass of substance `(m) xx` Specific heat `(c ) xx` Temperature change `(Delta T)` |
|
Answer» Accroding to `q = mc Delta T` `= (205 g) (4.18 J g^(-1) .^(@)C^(-1)) (91.4^(@)C - 21.2^(@)C)` |
|
| 2847. |
Give an example of an isolated system. |
|
Answer» Coffee held in a thermos flask is an isolated system because it can neither exchange energy nor matter with the surroundings. |
|
| 2848. |
What kind of system is the coffee held in a cup? |
|
Answer» The coffee held in a cup is an open system because it can exchange matter (water vapour) and energy (heat) with the surroundings. |
|
| 2849. |
What kind of system is the coffee held in a cup? |
|
Answer» Coffee held in a cup is an open system because it can exchange matter (water vapors) and energy (heat) with the surroundings. |
|
| 2850. |
State the first law of thermodynamics. |
|
Answer» The first law of thermodynamics stales that ‘the energy of an isolated system is constant’. |
|