Question

The water equivalent of a copper calorimeter is $4.5g$. If the specific heat of copper is $0.09\text{cal}{g}^{-1}{C}^{-1}$, then

• A. Mass of the calorimeter is $0.5\text{kg}$.
• B. Thermal capacity of the calorimeter is $4.5{\text{cal}}^{\circ }{C}^{-1}$.
• C. Heat required to raise the temperature of the calorimeter by $8^{\circ }C$ will be $36\text{cal}$
• D. Heat required to melt $15\text{gm}$ of ice placed in the calorimeter will be $1200\text{cal}$

Answer 1

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

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

$W=\left(\frac{ms}{S_{\omega }}\right)\Rightarrow 4.5=\frac{m\times 0.09}{1}$
$\therefore m=\frac{4.5}{0.09}=50g=0.05\text{kg}$
(b) Thermal capacity = ms
$=4.5\times 1{\text{cal}}^{\circ }{C}^{-1}=4.5{\text{cal}}^{\circ }{C}^{-1}$
$\left(c\right)Q=ms\Delta T=4.5\times 8\text{cal}=36\text{cal}$
$\left(d\right)Q=mL=15\times 80=1200\text{cal}$

Why this question? Key concept: Basic definition of water equivalent and thermal capacity. Importance in JEE: Questions related to basic concepts are common in JEE.