I need to solve latent heat. It gives me the number of ice cubes and the mass in grams (7550 grams). It wants to know how many calories would burn turning the ice into body temperature water (if body temp is 37 degrees celsius.) It gives Lm of H2O=1,44 kcal.mol and cwater=1.0 cal/g*degrees celsius. I can't figure out how to set up this problem.
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To solve this problem, we need to calculate the amount of heat energy required to raise the temperature of the ice cubes from their initial temperature (0 degrees Celsius) to the final temperature (37 degrees Celsius).
First, we need to determine the amount of heat energy required to heat the ice cubes from 0 degrees Celsius to 0 degrees Celsius (the phase change from solid to liquid). This is known as the latent heat of fusion. We can calculate this using the formula:
Q1 = mass of ice cubes * latent heat of fusion
The latent heat of fusion for water (Lm of H2O) is given as 1.44 kcal/mol. However, we need to convert it to calories per gram. Since the molar mass of water is approximately 18 g/mol, we can calculate the latent heat of fusion per gram as:
latent heat of fusion per gram = Lm of H2O / molar mass of water
Next, we can calculate the heat energy required to raise the temperature of the water from 0 degrees Celsius to 37 degrees Celsius. This can be calculated using the formula:
Q2 = mass of water * specific heat capacity * change in temperature
The specific heat capacity of water (cwater) is given as 1.0 cal/g*degrees Celsius.
Finally, the total heat energy required to convert the ice cubes to body temperature water would be:
Total heat energy required = Q1 + Q2
Now let's follow these steps and calculate the answer:
1. Calculate the latent heat of fusion per gram:
latent heat of fusion per gram = 1.44 kcal/mol / 18 g/mol = 0.08 kcal/g
2. Calculate Q1:
Q1 = mass of ice cubes * latent heat of fusion per gram = 7550 grams * 0.08 kcal/g = 604 kcal
3. Calculate Q2:
Q2 = mass of water * specific heat capacity * change in temperature
= 7550 grams * 1.0 cal/g*degrees Celsius * (37 - 0) degrees Celsius
= 279,850 cal
4. Calculate the total heat energy required:
Total heat energy required = Q1 + Q2
= 604 kcal + 279,850 cal (1 kcal = 1000 cal)
= 604 kcal + 279.85 kcal
= 883.85 kcal
Therefore, the number of calories required to turn the ice cubes into body temperature water is approximately 883.85 kcal.