5. The heat of vaporization of water is 2260 kJ/kg and the heat of melting of wa
ID: 1539269 • Letter: 5
Question
5. The heat of vaporization of water is 2260 kJ/kg and the heat of melting of water is 335 kJ/kg. The specific heat of water is 4186 J/(kg·°C)
a. How much energy will it take to melt a block of ice at 0oC with a mass 30 kg?
b. How much energy would it take to raise the temperature of the 30kg of water from 0oC to 100°C?
c. How much energy would it take to then completely convert the water to steam at 100oC?
d. If the energy used in heating for steps a, b and c above cost $0.08/KWh what is the cost for heating the ice to make steam?
Explanation / Answer
a) Energy required to melt the Ice, Q1 = m*Lf
= 30*335*10^3
= 1.005*10^7 J
b) Q2 = m*C*dT
= 30*4186*(100 - 0 )
= 1.256*10^7 J
c) Q3 = m*Lf
= 30*2260*10^3
= 6.780*10^7 J
d) Total eneergy, E = Q1 + Q2 + Q3
= (1.005 + 1.256 + 6.78)*10^7
= 9.04*10^7 J
= 9.04*10^7/(3.6*10^6) kWh
= 25.1 KWH
total cost = 25.1*0.08
= $2 <<<<<<<<<-------------------Answer
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