Calorimetry — Question 6

Given,

Mass of water (m) = 5 kg

Heat absorbed (Q) = 420 kJ = $420\times10^3$ J

Specific heat capacity of water (c) = 4200 J kg⁻¹ K⁻¹

Let, increase in temperature be △t.

From relation,

$Q = m c △t$

On rearranging terms,

$△t = \dfrac{Q}{mc}$

Substituting the values in the formula above we get,

$△t = \dfrac{420\times10^3}{5\times4200}=\dfrac{100}{5}=20 K= 20\ ^oC$

Temperature of water will increase by 20 °C.