Question

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

Answer 1

Let water equivalent to calorimeter be w.
Change in temperature = 5°C
Specific heat of water = 4200 J/Kg °C

Rate of flow of heat is given by

q = Energy per unit time = $\frac{ms∆T}{t}$

Case 1:

$q_1=\frac{\left(w+50\times {10}^{-3}\right)\times 4200\times 5}{10}$

Case 2

$q_2=\frac{\left(w+100\times {10}^{-3}\right)\times 4200\times 5}{18}$

From calorimeter theory, these two rates of flow of heat should be equal to each other.
$\Rightarrow q_1=q_2$

$\frac{\left(w+50\times {10}^{-3}\right)\times 4200\times 5}{10}=\frac{\left(w+100\times {10}^{-3}\right)\times 4200\times 5}{18}$
$\Rightarrow$18 (w + 50 × 10⁻³) = 10 (w + 100 × 10⁻³)
$\Rightarrow$w = 12.5 × 10⁻³ kg
$\Rightarrow$w = 12.5 g