Calorimetry — Question 5

Given,

Heat energy supplied (Q) = 1300 J

Mass of lead (m) = 0.5 kg

Change in temperature (△t) = (40 – 20)°C = 20° C

Specific heat capacity (c) = ?

From relation,

$Q = m \times c \times △t \\[0.5em]$

Substituting the values in the formula above we get,

$1300 = 0.5 \times c \times 20 \\[0.5em] \Rightarrow c = \dfrac{1300}{0.5 \times 20} \\[0.5em] \Rightarrow c = 130 \text{ J kg}^{-1} \text{K}^{-1}$

Hence, specific heat capacity of lead = 130 J kg^-1 K^-1