Question

1300 J of heat energy is supplied to raise the temperature of 0.5 kg of lead from ${20}^{0}C$ to ${40}^{0}C$. Calculate the specific heat capacity of lead.

• A. 100
• B. 130
• C. 150
• D. 200

Answer 1

The correct option is B

130

Heat energy, Q = 1300 J

Mass of lead, m = 0.5 kg

Rise in temperature, T = (40 – 20) = ${20}^{0}C$

Q = mc T

C = $\frac{Q}{mT}$ = $\frac{1300}{0.5\times 20}$ = 130 $Jk{g}^{-1}{}^0{C}^{-1}$​