Question

Part A:

How much heat is required to change a 37.9 g ice cube from ice at -14.1°C to water at 30°C? (if necessary, use cice=2090 J/kg°C and csteam= 2010 J/kg°C)

Part B:

How much heat is required to change a 37.9 g ice cube from ice at -14.1°C to steam at 112°C?

Answer 1

A)

ice at -14.1 to heated to ice at 0 degrees, ice at 0 degrees is heated to water at 0, water at 0 is heated to water at 30

Q = m*[ (cice*dT1) + Lf + (Cw*dT2) ]

Q = 0.0379*( (2030*14.1)+(334*10^3)+(4190*30))

Q = 18507.44 J

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ice at -14.1 to heated to ice at 0 degrees, ice at 0 degrees is heated to water at 0, water at 0 is heated to water at 100 , water at 100 oC to steam at 100 oC , stream at 100 oC to steam at 112 oC

Q1 = m*[ (cice*dT1) + Lf + (Cw*dT2) + Lv + (Csteam*dT3 ) ]

Q1 = 0.0379*( (2030*14.1)+(334*10^3)+(4190*100) + (2256*10^3) + (2030*12))

Q1 = 116049.2 J