# heat and energy

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Heat, Energy, and Changes of State

Part A

How much energy in joules would be absorbed by 3.0 L of liquid oxygen as it vaporized? The density of liquid oxygen at its boiling point is 1.14 kg/L , and its heat of vaporization is 213 kJ/kL .

Express your answer numerically in joules.

Part B

Some velomobile seats have been designed to hold ice packs inside their cushions. If you started a ride with ice packs that held 1400 g of frozen water at 0.0 degrees C , and the temperature of the water at the end of the ride was 32 C , how many calories of heat energy were absorbed?

Water's heat of fusion is 80 cal/g , and its specific heat is 1.0 .

Express your answer numerically in calories.

Part C

A canister is filled with 390 g of ice and 100 g of liquid water, both at 0.0 C. The canister is placed in an oven until all the H2O has boiled off and the canister is empty. How much energy in calories was absorbed?

Water's heat of fusion is 80 cal/g , its specific heat is 1.0 , and its heat of vaporization is 540 cal/g.

Express your answer numerically in calories.

Energy Conversions

Part A: Duncan has prepared ramen noodles so many times he does not need to measure the water carefully, but he knows that it takes 3.64Ã—104 to heat 1.00 of water from room temperature to boiling. If 0.534 of room-temperature water is heating on the stove, how many kilojoules of heat energy have been absorbed by the water at the moment it begins to boil?

Express your answer numerically in kilojoules.

Specific Heat Calculations

Part B: Duncan takes a break from studying and goes to the gym to swim laps. If he uses when he swims, how long will he have to swim to use 4.75Ã—105 ?

Part A: How much heat energy is required to raise the temperature of 0.363 of copper from 23.0 to 60.0 ? The specific heat of copper is 0.0920 .

Express your answer in calories.

Part B: If 125 of heat is applied to a 60.0- piece of copper at 25.0 , what will the final temperature be? The specific heat of copper is 0.0920 .

Express the final temperature numerically in degrees Celsius.

Problem 6.6

Part A: 3.1 kcal to cal

Part B: 525 J to kJ

Part C: 6550 cal to kJ

Part D: 6.50 kcal to J

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#### Solution Summary

The solution is composed to detailed explainations to heat and energy related questions.