Question

A calorimeter contains $50g$ of water at ${50}^{\circ }C$. The temperature falls to ${45}^{\circ }C$ in 10 min. When the calorimeter contains $100g$ of water at ${50}^{\circ }C$, it takes 15 min for the temperature to become ${45}^{\circ }C$. Find the water equivalent (in g) of the calorimeter. (Assume Newton’s law of cooling)

Answer 1

We know that: $Q=ms\Delta \theta$
Given:
$m=50g,{\theta }_1={50}^{\circ }C,{\theta }_2={45}^{\circ }C,$
$t_1=10min,m_2=100g,t_2=15min$

Let the water equivalent of calorimeter be $x$ gram.

Rate of heat loss is same for both cases
$\frac{dQ}{dt}=\frac{ms\Delta \theta }{t}$

$\frac{(50+x)s_w\times 5}{10}=\frac{(100+x)s_w\times 5}{15}$

$x=50g$