Atotal of 663 cal 663 cal of heat is added to 5.00 g 5.00 g of ice at − 20.0 °c . −20.0 °c. what is the final temperature of the water?

Final Temperature = 305.9 K or 32.9°C

Explanation:

Information given;

Mass (m) = 5g

Amount of Heat (H) = 663 calories

Initial Temperature = -20°C + 273 = 253K (Converting to Kelvin Temperature)

Final Temperature?

Specific Heat capacity of Water (c) =?

One might probably rush to use H = m * c * (T2-T1) to calculate the fianl temperature. That would have been wrong because Ice would melt when heat is being applied to it. And using that formular directly would lead to not considering the amount of heat required to melt it.

First let us check the amount of heat required to raise the temperature of the ice to 0°C

In this case now; Final temperature = 0°C + 273 = 273 K (Converting to Kelvin)

H = m * c * (T2-T1)

H = 5 * 1 * (273 - 253)

H = 5 * 1 * 20 = 100 cal

This shows the heat supplied is enough (663 cal is more than 100 cal) to bring the ice to its melting point.

Let's see if it would be sufficient to melt it.

Amount of Heat required to Melt ice;

H = mL;

where L = heat of fusion = 79.7 cal/g

H = 5 * 79.7 = 398.5 cal

Again, the heat is sufficient to melt it; the remaining heat would be used in raising the temperature of the liquid water.

In this case, initial temperature = 0°C + 273 = 273 K (Converting to Kelvin)

Amount of Heat left = 663 - 398.5 - 100 = 164.5 cal

Final temperature is given as;

H = m * c * (T2-T1)

164.5 = 5 * 1 * (T2 - 273)

T2 - 273 = 164.5 / 5

T2 = 273 + 32.9 = 305.9 K

Final Temperature = 305.9 K or 32.9°C

what are the two models

explanation: