Question

The water equivalent of a copper calorimeter is $4.5$ g. If specific heat of copper is $0.09cal/g{/}^{0}C$, then ($L_f=80cal/gm$)

• A. Mass of the calorimeter is $0.5$ kg
• B. Thermal capacity of the calorimeter is $4.5ca{l}^0/C$
• C. The heat required to raise the temperature of the calorimeter by $8^{0}C$ will be $36$ cal.
• D. Heat required to melt $15$ gm of ice at $0^{o}C$ Placed in the calorimeter will be $1200$ cal

Answer 1

Answer: (B), (C)

Thermal capacity of the calorimeter is $4.5ca{l}^0/C$
The heat required to raise the temperature of the calorimeter by $8^{0}C$ will be $36$ cal.

$\left(A\right)$ $ms=4.5$

$m\cdot \left(0.09\right)=4.5$

$m\left(\frac{9}{100}\right)=\frac{45}{10}$

$m=50gm=0.5kg$

$\left(B\right)$water equivalent $=\frac{ms}{S_w}$

Heat capacity$=4.5\times 1=4.5cal{/}^{0}C$

$\left(C\right)$ $Q=MS\Delta T$

$=4.5\times 1\times 8$

$=36cal$

Hence,

option $\left(B\right)$ and $\left(C\right)$ are both correct answer.