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

Explanation:

Step 1:

Let water be calorimeter-equivalent w.

Temperature-change = 5°C

water  of Specific heat = $4200 \mathrm{J} / \mathrm{Kg}^{\circ} \mathrm{C}$

Rate of heat flow is given by

$q=\frac{m s \Delta T}{t}$

Step 2:

first case

$q_{1}=\frac{(w+50 \times 10-3) \times 4200 \times 5}{10}$

second case

$q_{2}=\frac{(w+50 \times 10-3) \times 4200 \times 5}{18}$

From the theory of calorimeters these two heat flow rates should be equal to one another.

$q_{1}=q_{2}$

$\frac{(w+50 \times 10-3) \times 4200 \times 5}{10}=\frac{(w+50 \times 10-3) \times 4200 \times 5}{18}$

$18\left(w+50 \times 10^{-3}\right)=10\left(w+100 \times 10^{-3}\right)$

$w=12.5 \times 10^{-3} \mathrm{kg}$

$w=12.5 g$