Question

An electrically heating coil is placed in a calorimeter containing 360 g of water at $10\circ C$. The coil consumes energy at the rate of 90 W. The water equivalent of a calorimeter and the coil is 40 g. The temperature of water after 10 minutes will be:
$\text{A.}\quad 42.15^\circ C$
$\text{B.}\quad 32.14^\circ C$
$\text{C.}\quad 22.14^\circ C$
$\text{D.}\quad 52.14^\circ C$

Answer 1

A calorimeter works in such a way that when water comes inside its body, it provides some energy to the water, as per its power. If we know the rate at which heat is being provided to the water, we can directly compute the rise in heat and hence the rise in temperature.

Formula used:
$Q=ms\Delta T$, where ‘$s=4200 Jkg^{-1}K^{-1}$’ is the specific heat of water.

Complete step-by-step answer:
First of all, we need to know the concept of water equivalent. Every container has its own specific heat because it is also made up of some material. Hence we have to consider the heat which goes to raise the temperature of the container. Practically, it’s impossible that all 100% heat will go to raise the temperature of water. Now, water equivalent may be defined as the mass of container whose temperature will rise by the same value on giving some heat as risen of water by the same heat. The definition seems confusing but the trick to avoid the confusion is to take the mass of water added with the mass of container when calculating heat using $Q=ms\Delta T$
Now, the question is simple:
Given $m = 360 + 40 = 400\ g = 0.4\ kg$
Let the final temperature be ‘T’, so $\Delta T = T - 10$
Power of heater $= \dfrac{dQ}{dt} = 90 W$
Hence heat given by the heater to the water in 10 min = 600 sec = $90\times 600 = 54000\ J$
Now this heat will increase the temperature by relation:
$Q=ms\Delta T$
$54000 = 0.4\times 4200\times (T - 10)$
$T-10=32.14$
Or $T = 42.14^\circ C$

So, the correct answer is “Option A”.

Note: We don’t need to go for the derivation of water equivalent. Students are advised to just understand its significance and how to apply it in the problems. Also in many questions, the specific heat of water is not given. Students are advised to memorize the value of specific heat for water i.e. $s=4200 Jkg^{-1}K^{-1}$.