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A sphere of aluminium of 0.047 kg placed for sufficient time in a vessel containing boiling water, so that sphere is at 100" "^(@)C. It is then immediately transferred to 0.14 kg copper calorimeter containing 0.25 kg of water at 20" "^(@)C. The temperature of water rises and attains a steady state at 23" "^(@)C. Calculate the specific heat capacity of aluminium. |
| Answer» <html><body><p></p>Solution :Mass of sphere `m_(1)=0.047` <a href="https://interviewquestions.tuteehub.com/tag/kg-1063886" style="font-weight:bold;" target="_blank" title="Click to know more about KG">KG</a> <br/> Initial temperature of sphere `T_(1)=<a href="https://interviewquestions.tuteehub.com/tag/100-263808" style="font-weight:bold;" target="_blank" title="Click to know more about 100">100</a>^(@)<a href="https://interviewquestions.tuteehub.com/tag/c-7168" style="font-weight:bold;" target="_blank" title="Click to know more about C">C</a>` <br/> Final temperature of sphere `T_(2)=23^(@)C` <br/> Change in temperature `DeltaT_(1)=T_(1)-T_(2)` <br/> `=(100-23)" "^(@)C` <br/> `=77^(@)C` <br/> Sphere specific heat of sphere `=S_(1)` <br/> Heat lost by aluminum sphere, <br/> `Q_(1)=m_(1)S_(1)DeltaT_(1)` <br/> `=0.047xxS_(1)xx77` <br/> `:.Q_(1)=3.619S_(1)`. . .(1) <br/> Now, mass of water `m_(2)=0.25` kg <br/> Mass of calorimeter `m_(3)=0.14` kg <br/> Specific heat of water, <br/> `S_(2)=4.18xx10^(3)" Jkg"^(-1)K^(-1)` <br/> Specific heat of copper calorimeter, `S_(3)=0.386xx10^(3)" Jkg"^(-1)K^(-1)` <br/> Heat gained by water and calorimeter,<br/> `Q_(2)=m_(2)S_(2)DeltaT_(2)+m_(3)S_(3)DeltaT_(2)` <br/> `=0.25xx4.18xx10^(3)xx3+0.14xx0.386xx10^(3)xx3` <br/> `=3.135xx10^(3)+0.16212xx10^(3)` <br/> `:.Q_(2)=3.29712xx10^(3)`. . .(2) <br/> For <a href="https://interviewquestions.tuteehub.com/tag/steady-1226685" style="font-weight:bold;" target="_blank" title="Click to know more about STEADY">STEADY</a> state, <br/> {Heat lost by sphere} = {Heat gained by water} + {Heat gained by calorimeter} <br/> `Q_(1)=Q_(2)` <br/> `3.619S_(1)=3.29712xx10^(3)` [from equation (1) and (2)] <br/> `:.S_(1)=(3.29712xx10^(3))/(3.619)=0.911xx10^(3)` <br/> `:.S_(1)=0.<a href="https://interviewquestions.tuteehub.com/tag/911-342047" style="font-weight:bold;" target="_blank" title="Click to know more about 911">911</a>" kJ kg"^(-1)=911" Jkg"^(-1)K^(-1)` (specific heat of aluminium)</body></html> | |