Question

A 2.40 kg block of ice is heated with 5820 J of heat. The specific heat of water is 4.18 J•g^-1•C^-1. By how much will it’s temperature rise, assuming it does not melt?

Answer 1

Answer: The temperature rise is $0.53^0C$

Explanation:

The quantity of heat required to raise the temperature of a substance by one degree Celsius is called the specific heat capacity.

$Q=m\times c\times \Delta T$

Q = Heat absorbed by ice = 5280 J

m = mass of ice = 2.40 kg = 2400 g   (1kg=1000g)

c = heat capacity of water = $4.18J/g^0C$

Initial temperature  = $T_i$

Final temperature = $T_f$  

Change in temperature ,$\Delta T=T_f-T_i=?$

Putting in the values, we get:

$5280J=2400g\times 4.18J/g^0C\times \Delta T$

$\Delta T=0.53^0C$

Thus the temperature rise is $0.53^0C$