How much energy is required to change a 34 g ice cube from ice at -13?C to steam

Question

How much energy is required to change a 34 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 × 10^5 J/kg, and the heat of vaporization is 2.26 × 10^6 J/kg.
Answer in units of J

A 41 g ice cube at 0?C is placed in 570 g of water at 43?C. What is the final temperature of the mixture? The specific heat of water is 4186 J/kg · ?C and its latent heat of fusion is 3.33 × 10^5 J/kg .
Answer in units of ?C

Explanation / Answer

formula for calculating energy required, E = (heat energy required for ice to go from -13ºC to 0ºC) + (latent heat energy for ice at 0ºC to convert to water at 0ºC) + (heat energy for water to go from 0ºC to 100ºC) + (latent heat energy for water at 100ºC to convert to steam at 100ºC) + (heat energy for steam at 100ºC to steam at 111ºC) formula for heat energy required to change in temperature while in same state (for e.g., ice at -13ºC to ice at 0ºC) = m*c*(change in temperature) m = 42g = 0.042 hg c for ice 2.108 kJ/kg-K c for water 4.187 kj/kg-K c for steam 1.996 kJ/-kgK formula for latent heat energy = m*L L for ice to water = 337kJ/kg L for water to steam 2258kJ/kg Putting in the values you get, E = 0.042*2.108*13 + 0.042*337 + 0.042*4.187*100 + 0.042*2258 + 0.042*1.996*11 E = 128.64852 kJ

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