Question

A can of drinking chocolate contains a chemical heating pack that releases 3.5 x 104
J of
energy into drink when activated. Calculate the final temperature of the drink assuming no
energy is lost to the can or surroundings. [Mass of drink = 0.21 kg, specific heat capacity
of drink = 4200 J kg-1 K
-1
, initial temperature of drink = 20 °C]

Answer 1

The released energy is $Q$ , c is specific heat capacity of drink. The balance of energy is

$Q=cm (t_2-t_1), \;\; t_2 = t_1 + \dfrac{Q}{cm} = 20 °\text{C} + \dfrac{3.5\cdot10^4\,\mathrm{J}}{4200 \text{J\, kg}^{-1}\text{K}^{-1}\cdot0.21\,\mathrm{kg} } \approx 20 °\text{C} +39.7 °\text{C} = 59.7 °\text{C}.$