Question

Determine the time taken by a 500 W heater to raise the temperature of 50 kg of material of specific heat capacity $920\mathrm{J}.{\mathrm{kg}}^{-1}.{\mathrm{K}}^{-1}$, from ${18}^{\circ }\mathrm{C}$ to ${38}^{\circ }\mathrm{C}$. Assume that all the heat supplied by the heater is given to the material.

Answer 1

Step 1: Explanation of the given quantities:

Specific heat capacity $\mathrm{s}=920\mathrm{J}.{\mathrm{kg}}^{-1}.{\mathrm{K}}^{-1}$

Change in temperature$∆\mathrm{t}=\left({38}^{\circ }\mathrm{C}-{18}^{\circ }\mathrm{C}\right)={20}^{\circ }\mathrm{C}$

Power of the heater, $\mathrm{P}=500\mathrm{W}$

Mass of the material, $\mathrm{m}=50\mathrm{kg}$

Step 2: Calculation of the heat energy

$$\begin{gathered} \mathrm{Heat}\mathrm{energy}=\mathrm{mass}\times \mathrm{specific}\mathrm{heat}\mathrm{capacity}\times \mathrm{temperature}\mathrm{difference} \\ ∆Q=\mathrm{m}\times \mathrm{s}\times ∆\mathrm{t} \\ =50\mathrm{kg}\times 920\mathrm{J}.{\mathrm{kg}}^{-1}.{\mathrm{K}}^{-1}\times {20}^{\circ }\mathrm{C} \\ =920000\mathrm{J} \end{gathered}$$

Step 3: Calculation of the time

$$\begin{gathered} \mathrm{Time}=\frac{\mathrm{Heat}\mathrm{energy}}{\mathrm{power}} \\ \mathrm{t}=\frac{920000\mathrm{J}}{500\mathrm{W}} \\ \mathrm{t}=1840\mathrm{s} \\ \mathrm{t}=\frac{1840}{60} \\ \mathrm{t}=30\min 40\sec \end{gathered}$$

Step 4: Final answer

Thus the time taken by a 500 W heater to raise the temperature of a material is 30 min 40sec.