How many calories of heat are required to convert 220g of ice at -30°C to steam?
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How many calories of heat are required to convert 220g of ice at -30°C to steam?
How many calories of heat are required to convert 220g of ice at -30°C to steam?
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Answer:
The question is asking for the amount of heat (in calories) required to convert 220g of ice at -30°C to steam. The process of converting ice to steam involves three stages: melting, heating, and vaporization.
First, the ice must be melted, which requires heat called the heat of fusion. The heat needed to melt ice is 80 calories per gram. Therefore, to melt 220g of ice, 220 x 80 = 17600 calories of heat are needed.
Second, the melted ice must be heated, which requires the heat called the specific heat capacity. The specific heat capacity of water is 1 calorie per gram per degree Celsius. Therefore, to heat 220g of water from -30°C to 100°C (boiling point), we need to increase the temperature by 130°C. So, 220g x 130°C x 1 calorie/g°C = 28600 calories of heat are needed.
Finally, the water must be vaporized, which requires heat called the heat of vaporization. The heat needed to vaporize water is 540 calories per gram. Therefore, to convert 220g of water to steam, 220 x 540 = 118800 calories of heat are needed.
To sum up, the total amount of heat required to convert 220g of ice at -30°C to steam is the sum of the heat needed for melting, heating, and vaporization, which is:
17600 + 28600 + 118800 = 165000 calories of heat.