Question

A calorimeter contains $$70.2g$$of water at $${15.3}^{0}C$$. If $$143.7g$$ of water at $${36.5}^{0}C$$ is mixed with the common temperature is $${28.7}^{0}C$$. The water equivalent of the calorimeter is
(A) $$15.6g$$
(B) $$9.4g$$
(C) $$6.3g$$
(D) $$13.4g$$

Answer 1

Hint:-
Use the equation heat $$H=mC\mathrm{\Delta }T$$, $$m$$ is mass of the substance, $$C$$ is the specific heat and $$\mathrm{\Delta }T$$ is the change in temperature. Water equivalent means the mass of water which can raise the same amount of temperature in this process.

Complete step by step solution:-
In this process the heat is transferred from hot to colder body.
Total heat transferred is common heat transferred in a calorimeter.
Heat is given by $$H=mC\mathrm{\Delta }T$$
$$mC\mathrm{\Delta }T=m_{1}C\mathrm{\Delta }{T}_1+m_{2}C\mathrm{\Delta }{T}_2$$
But $$C$$is specific heat is common in all, everything is water in this problem.
Value of specific heat of water $$C=1calorie/gram$$
We can cancel out the specific heat.
$$m\mathrm{\Delta }T=m_1\mathrm{\Delta }{T}_1+m_2\mathrm{\Delta }{T}_2$$
Calorimeter contain water $$m_1=70.2g$$
Temperature of calorimeter $$T_1={15.3}^{0}C$$
Amount of water added is $$m_2=143.7g$$
Temperature of added water $$T_2={36.5}^{0}C$$
Common temperature or the equilibrium temperature $$T={28.7}^{0}C$$
Change in common temperature $$\mathrm{\Delta }T={28.7}^{0}C-0^{0}C={28.7}^{0}C$$
Change in calorimeter temperature $$\mathrm{\Delta }{T}_1=T_1-T={15.3}^{0}C-{28.7}^{0}C=-{13.4}^{0}C$$
Change in added water temperature $$\mathrm{\Delta }{T}_2=T_2-T={36.5}^{0}C-{28.7}^{0}C={7.8}^{0}C$$
$$m\mathrm{\Delta }T=m_1\mathrm{\Delta }{T}_1+m_2\mathrm{\Delta }{T}_2$$
Rearrange the equation
Water equivalent of calorimeter m
$$m={\displaystyle \frac{(m_1\mathrm{\Delta }{T}_1+m_2\mathrm{\Delta }{T}_2)}{\mathrm{\Delta }T}}$$
$$m={\displaystyle \frac{(70.2\times (-13.4))+(143\times 7.8)}{28.7}}$$
$$m={\displaystyle \frac{1120.86-940.68}{28.7}}={\displaystyle \frac{180.18}{28.7}}=6.27804878g$$
But in the option there are only two significant figures, and there only one digit after decimal point.
$$6.27804878g$$ Approximate to $$6.3g$$
So the answer is (C) $$6.3g$$

Note:- In this question we don’t need the value of Specific heat of water, it cancels outs in both sides of the equation. Calorimeter is the measuring device of heat produced in any kind of reaction which produces any kind of reaction and the material properties like heat capacity, etc.