Question

A 45.0-kg sample of ice is at 0.00°C. How much heat is needed to melt it? For water, Lf=334 kJ/kg and Lv=2257 kJ/kg 

A.  4.10 x 10^6 kj
B.  0.00 kj
C. 1.02 x 10^5 kj
D. 1.50 x 10 ^4 kj

Answer 1

H=ml
H=334×45.0
Heat Energy= 15030kJ

or 1.50×10^4kJ

answer: D

Answer 2

Heat required to change the phase of ice is given by

Q = m* L

here

m = mass of ice

L = latent heat of fusion

now we have

m = 45 kg

L = 334 KJ/kg

now by using above formula

`Q = 45 * 334 * 10^3`

`Q = 1.5 * 10^7 J`

In KJ we can convert this as

`Q = 1.5 * 10^4 kJ`

so the correct answer is D option