Question

An electric heater raises the temperature of 500 g of a given liquid from 25 ^OC to 31 ^OC in 120 seconds. If the power of heater is 1 kW, Calculate:

Specific Heat capacity of the liquid.

Answer 1

Step 1- Given Data:

Mass of liquid, $m=500g=\frac{500}{1000}Kg=0.5kg.$

Change in temperature, $∆T=31-25=6^{O}C$

Time,$t=120seconds.$

Power of heater, $P=1kW=1000W$

Step 2- To find out the heat capacity of the liquid i.e. C:

The formula for heat capacity is$s=\frac{P.t}{m.∆T}$.

$$\begin{gathered} \Rightarrow s=\frac{P.t}{m.∆T} \\ \Rightarrow s=\frac{\left(1000\right)\times \left(120\right)}{0.5\times 6} \\ \Rightarrow s=40000J/kg{-}^{O}C \end{gathered}$$

Thus, the heat capacity of the liquid is $s=40000J/kg-°C$.