Question

If a freezer cools 200 g of water from 20 ºC to its freezing point in 10 minutes, how much heat is removed per minute from the water?

Answer 1

Answer on Question 68285, Physics / Molecular Physics | Thermodynamics

Question:

If a freezer cools $200\,g$ of water from $20{}^{\circ}\mathrm{C}$ to its freezing point in 10 minutes, how much heat is removed per minute from the water?

Solution:

We can find how much heat is removed from the water during the freezing from the formula:

$$Q = mc\Delta t,$$

here, $m = 0.2\,kg$ is the mass of the water, $c = 4200\,J/kg\cdot{}^{\circ}\mathrm{C}$ is the specific heat capacity of the water and $\Delta t$ is the change in the temperature.

Then, we get:

$$Q = mc\Delta t = 0.2\,kg \cdot 4200\, \frac{J}{kg \cdot {}{}^{\circ}\mathrm{C}} \cdot (20{}^{\circ}\mathrm{C} - 0{}^{\circ}\mathrm{C}) = 16800\,J \approx 17000\,J.$$

Finally, we can find how much heat is removed per minute from the water:

$$Q_{per\ minute} = \frac{Q}{10\ min} = \frac{17000\,J}{10\ min} = 1700\, \frac{J}{\min}.$$

Answer:

$$Q_{per\ minute} = 1700\, \frac{J}{\min}.$$

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