Question

4. A liquid of specific heat capacity of 3000 JKg-1K
-1
rise from 15 oC to 65 oC in one minute
when an electric heater is used. If the heater generates 63000 J per minutes, calculate
the mass of the water.

Answer 1

The heat $Q$ generated in one minute is:

$Q = 63000(J/min)·1(min) = 63000 (J)$.

Assuming that no heat was lost , the heat generated equals the heat absorbed by the liquid.

The relation between the heat $Q$ absorbed by the liquid and the temperature change $T_2-T_1$ is the following:

$Q = cm(T_2-T_1)$ ,

where $c$ is the specific heat capacity of the liquid and $m$ is its mass. The temperature change is the difference temperature and has the same numerical value in °C and in K, therefore $T_2-T_1 = 65-15 = 50$ K.

Making use of this relation, we can deduce the mass of the liquid:

$m = \frac{Q}{c(T_2-T_1)} = \frac{63000 (J)}{3000(Jkg^{-1}K^{-1})·(50) (K)} = 0.42$ kg.

Answer: the mass of the water is 0.42 kg.