A 2100W continuous flow geyser has water inlet temperature of $10^{\circ}C$ while the water flows out at a rate of $20gs^{-1}$. The outlet temperature of the water must be:
$\text{A}. \ 20^{\circ}$
$\text{B}. \ 30^{\circ}$
$\text{C}. \ 35^{\circ}$
$\text{D}. \ 40^{\circ}$

Answer Verified Verified
Hint: A geyser works in such a way that when water comes inside the body of geyser, it provides some energy to the water, as per its wattage. After some time of heating, when hot water is required by someone, the water is taken out of the geyser.

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

Complete answer:
Given, $\dfrac{dQ}{dt}=2100\ W \ and \ \dfrac{dm}{dt}=20gs^{-1}=0.02kgs^{-1}$.
In order to relate the given values, we must differentiate the expression $Q=ms\Delta T$ with respect to time.

i.e. $\dfrac{dQ}{dt} = \dfrac{dm}{dt}s\Delta T$

On putting the values, we get

$2100 = 0.02\times 4200 \Delta T$
Or $\Delta T = \dfrac{2100}{0.02\times 4200}= 25$
As $\Delta T = T_f-T_i$
Hence $T_f-10=25$
Or $T_f=35^{\circ}$
Hence the outlet temperature is $35^{\circ}$.

So, the correct answer is “Option C”.

Additional Information:
Generally the specific heat of the fluid involved in the question is given, but in some cases, standard fluids like water is used. Hence we should have an idea of their values, for example, for water, $s=4200Jkg^{-1}K^{-1} or \ 1 \ cal \ g^{-1}K^{-1}$ which generally means that 1kg of water require 4200J of energy to raise the temperature by $1^{\circ}C \ or \ 1K$. Also 1 g of water requires 1cal of heat to raise its temperature by $1^{\circ}C \ or \ 1K$. 1 calorie and 1 Joule relates as 1 calorie = 4.2 Joule.

In the question, we are required to get the relation between rate of flow of heat and rate of flow of water. This could only be done by differentiating the expression in which both heat and mass terms appear. We can also come across many questions in which specific heat varies with temperature. But for smaller changes in temperature, it is assumed to be constant. Such questions are only dealt using the above method of differentiation.