Question

How much energy is required to change a 44 g
ice cube from ice at −13◦C to steam at 111◦C?
The specific heat of ice is 2090 J/kg · ◦ C, the
specific heat of water is 4186 J/kg · ◦ C, the
specific heat of stream is 2010 J/kg · ◦ C, the
heat of fusion is 3.33 × 105
J/kg, and the heat
of vaporization is 2.26 × 106 J/kg.
Answer in units of J.

Answer 1

$$Q=Q_{melting}+Q_{vapor}+Q_{heat}^{ice}+Q_{heat}^{water}+Q_{heat}^{steam}$$

$$Q=(0.044)(3.33 \cdot10^5+2.26\cdot10^6+\\+(2090)(13)+(4186)(100)+(2010)(11))$$

$$Q=1.35\cdot 10^5J$$