Question

A calorimeter contains $50\text{g}$ of water at ${50}^{\circ }C$. The temperature falls to ${45}^{\circ }C$ in $10\text{minutes}$. When the calorimeter contains $100\text{g}$ of water at ${50}^{\circ }C$, it takes $18\text{minutes}$ for the temperature to reduce to ${45}^{\circ }C$. The water equivalent of the calorimeter is (in $\text{g}$)
[Answer up to two decimal places]

Answer 1

We know that rate of heat transfer

$H=\frac{d\theta }{dt}=\frac{-KAdT}{dx}$

Here for calorimeter, area of the calorimeter $\left(A\right)$, thickness $\left(dx\right)$ of wall of calorimeter and temperature difference $\left(dT\right)$ are constant.

Thus, $H=\text{constant}$

$\Rightarrow (\frac{dQ}{dt}{)}_1=(\frac{dQ}{dt}{)}_2.......\left(1\right)$

Let us suppose water equivalent of the calorimeter is $w$, then from eq. (1)

$\frac{(0.05+w)s\Delta T}{t_1}=\frac{(0.1+w)s\Delta T}{t_2}\because [\Delta Q=ms\Delta T]$

$\Rightarrow \frac{0.05+w}{10}=\frac{0.1+w}{18}$

$\Rightarrow 0.9+18w=1+10w$

$\Rightarrow 8w=0.1$

$\Rightarrow w=0.0125\text{kg}=12.5\text{g}$

Why this question? Concept involved :- The water equivalent of a body is equivalent to the amount of water that is required to consume the same quantity of heat as that substance does for a temperature rise of one degree. the water equivalent of a body is indeed equivalent to the product of its mass and its specific heat.