Question

A calorimeter contains 50 g of water at 50^0 C. The temperature falls to ${45}^0$ C in 10 minutes. When the calorimeter contains 100 g of water at ${50}^0$ C, it takes 18 minutes for the temperature to become ${45}^0$ C. Find the water equivalent of the calorimeter.

Answer 1

Let the water equivalent of calorimeter

= m

In 1st case,

$\frac{(m+50\times {10}^{-}3)\times 4200\times 5}{10}$

= rate of heat flow

In second case,
$\frac{(m+100\times {10}^{-3})\times 4200\times 5}{18}$

= rate of heat flow

From (i) and (ii),

$\frac{(m+50\times {10}^{-3})\times 4200\times 5}{10}$

= $\frac{(m+100\times {10}^{-3})\times 4200\times 5}{18}$

$\Rightarrow (m+50\times {10}^{-3})\times 18$

=$10m+1000\times {10}^{-3}$

$\Rightarrow 18m+18\times 50\times {10}^{-3}$

= $10m+1000\times {10}^{-3}$

$\Rightarrow 8m=100\times {10}^{-3}$ kg

m = $12.5\times {10}^{-3}$kg = 12.5 gms.