A 9.83 kg block of ice has a temperature of -15.5^{o} C. The pressure is one atmosphere. The block absorbs 4.404×10^{6} J of heat. What is the final temperature of the liquid water?

The heat capacity of ice at 1atm is 2.11J·g−1·K−1

So for 9.83 kg of ice to reach 0 C from -15.5 C,

the heat needed=9.83x10^3x15.5x2.11=3.21x10^5 J

The heat of fusion for melting ice is 334 J/g

So for 9.83kg of ice to change into water,

the heat needed=9.83x10^3x334=3.28x10^6 J

So far, the total heat used=3.28x10^6 + 3.21x10^5 =3.601x10^6 J

So the heat left=4.404×10^6-3.601x10^6 =0.803x10^6 J

The heat capacity of liquid water is 4.1813 J·g−1·K−1.

So the temperature change due to the left heat is

T-T0=0.803x10^6/4.1813/(9.83x10^3)=19.5 K

T=19.5k+T0=19.5+273 K=19.5+273-273 C=19.5 C

So the final temperature of the liquid water is 19.5C

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