Q138 / 160 · TS EAMCET 2020 (14 Sep Shift 1) · Physics
How much heat energy is supplied when $5 \mathrm{~kg}$ of water at $20^{\circ} \mathrm{C}$ is brought to its boiling point? (Assume, specific heat of water $=4.2 \mathrm{~J} / \mathrm{g}^{\circ} \mathrm{C}$ )
- Correct answerA1680 kJ
- B1740 kJ
- C1680 J
- D1740 J
Solution
Given that, $m=5 \mathrm{~kg}$
$\begin{gathered}
c=4.2 \frac{\mathrm{J}}{\mathrm{g} \times{ }^{\circ} \mathrm{C}}=4.2 \frac{\mathrm{J}}{10^{-3} \mathrm{~kg} \times{ }^{\circ} \mathrm{C}} \\
=4.2 \times 10^3 \frac{\mathrm{J}}{\mathrm{kg} \times{ }^{\circ} \mathrm{C}}=4200 \frac{\mathrm{J}}{\mathrm{kg} \times{ }^{\circ} \mathrm{C}} \\
T_1=20^{\circ} \mathrm{C}, T_2=100^{\circ} \mathrm{C}, \Delta Q=?
\end{gathered}$
Since, $\begin{aligned} \Delta Q & =m c \Delta T=m c\left(T_2-T_1\right)=5 \times 4200(100-20) \\ & =5 \times 4200 \times 80=1680000 \mathrm{~J} \\ & =1680 \times 10^3=1680 \mathrm{~kJ}\end{aligned}$
Hence, the correct answer is option: 'A'.