Question

calculate the amount of heat required to change 10 g  of ice at 20^0 c to water at 40^0 c. Take specific heat capacity of ice = 2.1 j/g/^0c, specific latent heat of ice =336 j/g & specific heat capacity of water =4.2 j/g/^0c

Answer 1

THERE is a mistake in the given  DATA. ICE is at -20° C  and NOT at 20° C.  Minus twenty deg.

We assume that the Ice is first Heated slowly from -20 ° C to 0° C retaining the state as ICE.

Then we convert the ICE into water at 0 ° C.
       Amount of heat required to heat ICE at -20 ° C to ICE at 0°C : 
           = mass * specific heat * (T2-T1) = 10 gm * 2.1 J/gm/°C * (0°C - (-20° C))
                           = 420 Joules
     
To convert ice to water at a constant temperature 0C, energy required :                            = Latent Heat * mass = 336 J/ gm * 10 gms = 3360 J

Now water is heated steadily to 40 ° C from 0° C.
       Amount of heat required to heat water at 0° C to 40 °C : 
           = mass* sp heat * (T2-T1) = 10 gm * 4.2 J/gm/°C * (40°C - 0° C)
               = 1680 Joules

Total amount of heat needed = 420 J + 3360 J + 1680 J
                    = 5, 460 J    or 5.460 kJ