Heat is added to 1-99 kg of ice at -25 C- How many kilocalories are re.docx

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This document provides a step-by-step solution to calculate the number of kilocalories required to change 1.99 kg of ice at -25 degrees Celsius to steam at 144 degrees Celsius. It lists the relevant properties and formulas, and calculates that 1502.57 kilocalories are required based on the mass, specific heats, and latent heats of fusion and vaporization of water and ice over the temperature ranges.

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Heat is added to 1.99 kg of ice at -25 Â°C. How many kilocalories are required to change the ice
to steam at 144 Â°C?
If you could show me how to set this problem up it would be much appreciated! I've been
staring at it for a while and don't even know where to start. Thank you!
Solution
mass = m = 1.99 kg = 1990g
latent heat of fusion of ice = L1 = 80 cal / g
latent heat of vaporization of water = L2 = 540 cal / g
Specific heat of ice = C1 = 0.5 cal /g K
Specific heat capacity of water = C2 = 1 ca/g K
Specific heat capacity of steam = C3 = 0.48 cal/ g K
heat required = m ( C1(0 + 25) + L1 + C2(100-0) + L2 + C3(144-100)
= 1990 * ( 0.5*25 + 80 + 1*100 + 540 + 0.48*47)
= 1502569.4calories
= 1502.57 Kilocalories

Heat is added to 1-99 kg of ice at -25 C- How many kilocalories are re.docx

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