What quantity of heat is required for a 120g of ice at -20°C melting to water at 20°C?

 The complied data is

 

m = 120 g

Cp^sl = 2.1 J·g−1

Cp^l = 4.226 J⋅g−1⋅°C−1.

ΔH^sl = 333.5 J·g−1.

T^s = -20°C

T^o = 0°C

T^l = 20°C

 

The equation is

 

Q = m {[(Cp^l Cp^sl) x (T^o - T^s)] + ΔH^sl + Cp^l x (T^l - T^o)}

 

The equation with values is:

 

Q = 120 x {[(4.226 – 2.1) x (0+20)] + 333.5 + 4.226(20-0)}

 

The solution is

 

Q = 55254.8 J