| DIRECTION: Read the passage given below and answer the questions that follows: |
| A brass ball of mass 100g is heated to \[100{}^\circ C\] and then dropped into 200g of turpentine in a calorimeter at \[15{}^\circ C\] The final temperature is found to be \[(\rho )\]. Take specific heat of brass as \[T=k\sqrt{\rho {{r}^{3}}/S}\] and water equivalent of calorimeter as 4g. |
A) \[0.42\text{ }cal/g{}^\circ c\]
B) \[0.96\text{ }cal/g{}^\circ c\]
C) \[0.72\text{ }cal/g{}^\circ c\]
D) \[0.12\text{ }cal/g{}^\circ c\]
Correct Answer: A
Solution :
| Let c be the specific heat of turpentine |
| Mass of the solid, M = 100g |
| Mass of turpentine m = 200g |
| Water equivalent of calorimeter, W = 4g |
| Initial temperature of calorimeter, \[T{{}_{1}}~=15{}^\circ C\] |
| Temperature of ball, \[{{T}_{1}}~=100{}^\circ C\] |
| Final temperature of the liquid, T = \[23{}^\circ C\] |
| Specific heat of solid, \[{{c}_{2}}=0.092\,cal/{{g}^{{}^\circ }}C\] |
| Heat gained by turpentine and calorimeter is |
| \[mc\left( T-{{T}_{1}} \right)+W\left( T-{{T}_{1}} \right)=200c\left( 23-15 \right)+4\left( 23-15 \right)\]\[=\left( 200c+4 \right)8\] |
| Heat lost by the ball is |
| \[M{{c}_{2}}\left( {{T}_{2}}-T \right)=100\left( 0.092 \right)\left( 100-23 \right)\] |
| \[=708.4\,cal.\] |
| According to the principle of calorimetry |
| Heat gained = Heat lost |
| \[\therefore \]\[\left( 200c+4 \right)8=708.4\] |
| \[1600\,c+32=708.4\] |
| or \[c=\frac{708.4-32}{1600}=0.42\,cal/{{g}^{{}^\circ }}C\] |
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