Question

250 g of water and equal volume of alcohol of mass 200 g are replaced successively in the same calorimeter and cool from ${60^ \circ }$ to ${55^ \circ }$ in the 130s and 67s respectively. If the water equivalent of the calorimeter is 10 g, then the specific heat of alcohol in ${{cal} \mathord{\left/ {\vphantom {{cal} {gm}}} \right.} {gm}}{ \times ^ \circ }C$ is:
(A) $1.25$
(B) $0.69$
(C) $0.62$
(D) $0.68$

Answer 1

The temperature change per second is calculated for both water and alcohol. The principle of calorimetry is then used to solve the question.

Complete Step by Step Solution
Firstly, we have to calculate the temperature change per second during the process in the calorimetry.
For water, $\Delta {t_w} = \dfrac{{{\theta _1} - {\theta _2}}}{{{t_w}}}$ where ${\theta _1}$ is the initial temperature, ${\theta _2}$ is the final temperature and ${t_w}$ is the time taken by water for the change in temperature.
Given that, 250 g of water cools from ${60^ \circ }$ to ${55^ \circ }$ in 130s.
Thus, $\Delta {t_w} = \dfrac{{{{60}^ \circ } - {{55}^ \circ }}}{{130}} = {\dfrac{5}{{130}}^ \circ }{C \mathord{\left/ {\vphantom {C s}} \right.} s}$ .
For alcohol, $\Delta {t_a} = \dfrac{{{\theta _1} - {\theta _2}}}{{{t_a}}}$ where ${\theta _1}$ is the initial temperature, ${\theta _2}$ is the final temperature and ${t_a}$ is the time taken by alcohol for the change in temperature.
Given that, 200 g of alcohol, of the same volume as that of water cools from ${60^ \circ }$ to ${55^ \circ }$ in 67s.
Thus, $\Delta {t_w} = \dfrac{{{{60}^ \circ } - {{55}^ \circ }}}{{67}} = {\dfrac{5}{{67}}^ \circ }{C \mathord{\left/ {\vphantom {C s}} \right.} s}$ .
Now, by applying the principle of calorimetry, the total heat lost by the hot body is equal to the total heat gained by the cold body.
Hence, heat gained = heat lost.
So, $\left( {m + {m_1}} \right){s_1}\Delta {t_w} = {m_2}{s_2}\Delta {t_a}$ .
where $m$ is the mass of calorimeter, ${m_1}$ is the mass of water, ${m_2}$ ​is the mass of alcohol, ${s_1}$ ​is the specific heat of water and ${s_2}$ is the specific heat of alcohol.
In the question, it has been stated that the water equivalent of the calorimeter is 10 g.
Assigning the values, $m = 10g$ , ${m_1} = 250g$ , ${m_2} = 200g$ and ${s_1} = 1{{cal} \mathord{\left/ {\vphantom {{cal} g}} \right.} g}{ \times ^ \circ }C$ .
Thus, we may write, $\left( {250 + 10} \right) \times 1 \times \dfrac{5}{{130}} = 200 \times {s_2} \times \dfrac{5}{{67}}$ .
Simplifying, ${s_2} = 260 \times \dfrac{{67}}{{130}} \times \dfrac{1}{{200}}$
So, ${s_2} = 0.68{{cal} \mathord{\left/ {\vphantom {{cal} g}} \right.} g}{ \times ^ \circ }C$ .
Hence the correct answer is Option D.

Note
The act or science of measuring the changes in the state variables of a body in order to calculate the heat transfer associated with changes of its states such as physical changes or phase transitions under specific conditions is known as calorimetry. Calorimetry is performed with the help of a calorimeter.
When two bodies of different temperatures (preferably a solid and a liquid) are placed in physical contact with each other, the heat is transferred from the body with higher temperature to the body with lower temperature until thermal equilibrium is attained between them. The body at higher temperature releases heat while the body at lower temperature absorbs heat. The principle of calorimetry indicates the law of conservation energy, i.e. the total heat lost by the hot body is equal to the total heat gained by the cold body.