Question

A 78 kg block of Iron (Specific Heat Capacity 348 J/kg.K) at temperature T kelvin is inserted into 5 kg of water (Specific Heat Capacity 4005 J/kg.K and initial temperature is 300 K). Final temperature of the mixture is 308 K. Assuming no heat loss to surrounding calculate the initial temperature of the iron block: _______ K

Answer 1

mass of iron block = mb = 78 kg

initial temperature = T
                           = cb = 348 j/kg k
mass of water = mw = 5 kg
initial temperature of water = 300 k

from law of calorimetrey

heat lost by iron = heat gain by water

mb* cb* ( T- 308 ) = mw * cw * ( 308-300 )
78 * 348* ( T- 308 ) = 5 *4005* 8
27144 ( T- 308 ) =160200

           T - 308 = 5.9

             T = 313.9 = 314 k