Question

# 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}$

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.

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”.

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.