Question

The water equivalent of a copper calorimeter is $4.5\text{g}$. If the specific heat of copper is $0.09{\text{cal/g}}^{\circ }\text{C}$, choose the correct option(s):
[Latent heat of fusion of ice $=80\text{cal/g}$]

• A. Mass of the calorimeter is $0.5\text{kg}$.
• B. Thermal capacity of the calorimeter is $4.5{\text{cal/}}^{\circ }\text{C}$.
• C. Heat required to raise the temperature of the calorimeter by $8^{\circ }\text{C}$ will be $36\text{cal}$.
• D. Heat required to melt $15\text{g}$ of ice placed in the calorimeter will be $1200\text{cal}$.

Answer 1

Answer: (B), (C), (D)

Heat required to melt $15\text{g}$ of ice placed in the calorimeter will be $1200\text{cal}$.

Water equivalent is given by
$W=ms$
where $m\to$ mass of the substance
$s\to$ specific heat capacity
or $m=\frac{W}{s}=\frac{4.5}{0.09}=50\text{g}=0.05\text{kg}$

The thermal capacity and the water equivalent of a body have the same numerical value.
i.e Thermal capacity of copper $S=4.5{\text{cal/}}^{\circ }\text{C}$

Also, heat required to raise the temperature of the copper calorimeter by $8^{\circ }\text{C}$ will be
$Q=4.5\times 8=36\text{cal}$

Since the temprature remains constant during the process of melting, no heat is exchanged with the calorimeter and hence,
$Q=m_{ice}\times L_f=15\times 80=1200\text{cal}$

Hence, the correct choices are (b), (c) and (d).