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The telephone wires will sag in summer due to expansions and will break in winter due to contraction.

 Therefore, while putting up the wires between the poles, care is taken that in summer they are kept slightly loose so that they may not break in winter due to contraction. While in winter they are kept light so that they may not sag too much in summer due to expansion

Correct answer is (c) 50°

The reason is that the borosilicate glass do not expand much on being heated and therefore they do not crack.

Correct option is  D)50g

m = 5 kg = 5000 g The heat energy released to convert 5 kg of water at 40°C to 5 kg of water at 0°C.

Q₁ = ms∆T 

= 5000 × 1 × (40 – 0) 

= 5000 × 40 = 200000 cal [s = 1 cal/g-°C for water] 

The heat energy released to convert 5 kg of water at 0°C to 5 kg of ice at 0°C. 

Q₂ = mL_f

= 5000 × 80 = 400000 cal [∵ L_f = 80 cal/g] 

Total energy released = 200000 + 400000 

= 600000 cal 

= 600 kcal 

= 142.86 kJ.

Correct option is  A) 900 cal

The energy required to convert 1 g of ice at 0°C into lg of water at 0°C is latent heat of fusion that is 80 cal/g.

In CGS system :

Mass of water = m = 1 gm 

Latent heat of fusion of ice (L) = 80 cal/gm 

The energy transferred or released when 1 gm of water at 0° C freezes to ice at 0° C. 

Q = mL_freeze = 1 × 80 = 80 Cal.

(OR) 

In SI system : 

Mass of water = m = 1 gm = 1/1000 kg

Latent heat of fusion = L = 3.36 × 10⁵ J/kg.

Amount of heat released or transferred when lgm of water at 0°C freezes to ice at 0°C.

Q = mL_fusion = 1/1000 kg × 3.36 × 10⁵ = 3.36 × 10² = 336J.

(OR) 

In CGS system : 

Mass of water = m = 1 gm 

Latent heat of fusion of ice (L) = 80 cal/gm

The energy transferred or released when 1 gm of water at 0° C freezes to ice at 0°C. 

Q = mL_freeze = 1 × 80 – 80 cal. 

(Or) 

In SI system : 

In SI, Q = 80 × 4.2 [1 cal = 4.2 J] 

Q = 1 x 10^-3 × 3.36 × 10⁵ = 3 36 J.

Correct option is  A) 3, 500 cal

Correct option is  B) J/Kg

Correct option is  A) cal/gm

Correct option is  B) 80 cal

We will notice that the grains of food colour move randomly (jiggle). This happens because of the molecules of water on both bowls are in random motion.

Temperature in Kelvin = 273 + Temperature in degree Celsius. 

[K = t°C + 273]

i) Copper. Because, it has low specific heat value.

ii) Due to high specific heat value of water, it can take more time to heat itself and acts as a coolent.

1. We know that the temperature of the body is directly proportional to the average kinetic energy of particle of the body. 

2. The molecules of the system have different forms of energies such as linear, rotational kinetic energy, vibrational energy and potential energy. 

3. When we supply heat energy, it will be shared in different forms and increase the energy in the system. 

4. This sharing will vary from material to material. 

5. If the maximum share of heat energy is spent to rise linear kinetic energy, then the system gets increasing in temperature

6. Due to differences in sharing different materials have different specific heats.

1. We know that the temperature of a body is directly proportional to the average kinetic energy of particles of the body.

2. The molecules of the system have different forms of energies. The total energy of the system is called internal energy of the system.

3. When we supply heat energy to the system, the heat energy given to it will be shared by the molecules among the various forms of energy.

4. This sharing will vary from substance to substance.

5. The rise in temperature is high for a substance, if the maximum share of heat energy is utilised for its linear K.E.

6. This sharing of heat energy also varies with temperature. That is why the specific heat is different for different substances.

1. The substance ‘C’ has the highest specific heat. 

2. Because ‘C’ does not rise its temperature quickly. In other words ‘C’ takes more time to rise its temperature.

i) The melting point of the solid is T₁.

ii) The latent heat of fusion is (H₂ – H₁)

iii) The latent heat of vaporisation is (H₄ – H₃)

iv) The boiling point of the substance is T₃.

1. clinical 

2. mercury 

3. air pressure 

4. climate 

5. humidity

C) 540 cal/gm

Correct option is  B) J/Kg

 Q = mL

so, L = Q/m.

Hence, the correct option is (B) Q/m

Correct option is  B) Q/m

Data: m₁ = 100 g, L₁ = 540 cal/g,

T₁ = 100 °C, mass of ice, m = ?, L₂ = 80 cal/g, c

(water) = 1 cal/g·°C

According to the principle of heat exchange, heat lost by hot body = heat gained by cold body. Conversion of steam into water:

Q₁ = m₁L₁ = 100 g × 540 cal/g = 54000 cal

Decrease in the temperature of this water to 0 °C:

Q₂ = m₁c × (T₁ – 0 °C) = 100 g × 1 cal/g·°C × (100 °C – 0 °C) = 10000 Cal

Melting of ice: Q₃ = mL₂

= m × 80 cal/g

Now, Q₁ + Q₂ = Q₃

∴ (54000 + 10000) cal = m × 80 cal/g

∴ m = \(\frac{64000}{80}\) = 800 g

800 g of ice will melt.

1. Clinical Thermometer is used in hospitals to measure the temperature of the human body. 

2. It has a kink that prevents the mercury from flowing back into the bulb when it was taken out of the patient’s mouth. 

3. This helps to record the temperature conveniently.

Clinical thermometer 

1. Wash the clinical thermometer properly with an antiseptic solution. 

2. To lower the mercury level, hold the thermometer firmly and give some jerks. 

3. Ensure that it falls below 35°C. Now place the bulb of the thermometer under your tongue. 

4. After one or two minutes, take the thermometer out and note the reading. 

5. This is your body temperature. 

6. Don’t hold the thermometer by the bulb while reading it. 

A Laboratory thermometer: 

1. Take cold or hot water in a bowl. Place the mercury bulb of the thermometer in the water so that the bulb immerses in it. 

2. Wait for some time till the mercury level shows a constant reading. 

3. Note down that reading.

We cannot measure too low and too high temperatures by using a clinical thermometer. It is made for clinical use only, so by using this we can measure from 35°C to 42°C only. If we measure temperature of boiling water or ice cubes it will break or it does not work.

(b) not flow from iron ball to water or from water to iron ball.

Specific heat capacity, C = \(\frac{Q}{mx \Delta T}\)

Here, m = 100 kg

Heat capacity = \(\frac{Q}{\Delta T}\)  = 8000 J/°C = 8000 J/K

C = \(\frac{Q}{mx \Delta T}\) = 100 x 8000 J = 8,00,000 JKg^-1 K^-1

The correct option is B) Digital Thermometer.