Question

How much energy is required to heat a frozen can of juice (360 grams- mostly water) from 0 degrees Celsius ( the temperature of an overcooled refrigerator) to 110 degrees ( the highest practical temperature within a microwave oven)?

Answer 1

The energy required to heat the frozen can of juice from 0 degrees Celsius to 110 degrees Celsius is 163.08 kJ.

Step-by-step explanation:

To calculate the amount of energy required to heat the frozen can of juice, we need to consider three steps:

Heating the can from 0 degrees Celsius to its melting point, which is 0 degrees Celsius

Melting the ice at 0 degrees Celsius

Heating the water from 0 degrees Celsius to 110 degrees Celsius

The energy required for each step can be calculated using the formula Q = mcΔT, where Q is the energy, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.

For the first step, Q = (360g)(4.18 J/g°C)(0 - 0) = 0 J.

For the second step, Q = (360g)(333.55 J/g)(0 - 0) = 0 J.

For the third step, Q = (360g)(4.18 J/g°C)(110 - 0) = 163,080 J = 163.08 kJ.

Adding up the energies from each step, the total energy required to heat the frozen can of juice is 163.08 kJ.

Answer 2

1,100,160J or 262.94 kcal

Step-by-step explanation:

The juice is frozen at 0 degrees Celsius and I assume that it will become gas at 100 degrees Celsius. So we change the form of the water from solid to liquid, then to gas. That means we have to find out how much heat needed to change water form too, not only the heat needed to increase its temperature.

The latent heat of water is 4.2J/g °C while the heat of fusion is 334 J/g and the heat of vaporization is 2260 J/g. The energy needed will be:

360g * 4.2J/g °C * (110-0°C ) + 360g * 334 J/g + 360g * 2260 /g = 1,100,160J or 262.94 kcal.