Calculate the quantity of heat needed to change the temperature of 60g of ice at 0 ^\circC to 80 ^\circC. [specific latent heat of fusion of ice= 3.36 x 10\(^5\) Jkg\(^{-1}\) specific heat capacity of water 4.2 x 10\(^3\) J kg\(^{-1}\) K\(^{-1}\)]
A
4.80 kJ
B
20.16 kJ
C
40.32 kJ
D
22.17 kJ
Verified Correct Answer: C.
Physical Principles & Numerical Solution
Heat needed to change 60g (0.06kg) of ice at 0\(^o\)C to water at 0\(^o\)
= 0.06 x 3.36 x 10\(^5\)J
= 0.2016 x 10\(^5\) J
Heat required to raise the temp. of 60g (0.06kg) of water from 0\(^o\)C to 80\(^o\)C
= 0.06 x 4.2 x 10\(^3\) x (80.0)J
= 0.06 x 4.2 x 10\(^3\) x 80
= 20.16 x 10\(^3\)
= 0.2016 x 10\(^5\)
Total heat needed = (0.2016 + 0.2016) 10\(^5\)J
= 0.4032 x 10\(^5\) J
= 40.32KJ
Source: Myschool Past Questions Archive (Record #58857)