how long would a 2.68 kw space heater have to run to put into a kitchen the same amount of heat as a refrigerator (coeff of performance = 2.13) does when it ffreezes 1.71 kg of water at 20 into ice at 0

To solve this problem, we need to calculate the amount of heat removed by the refrigerator (when freezing water), and then determine how long the space heater would need to run to produce the same amount of heat.

### Step 1: Calculate the heat removed by the refrigerator

The refrigerator removes heat from the water as it cools from 20°C to 0°C, then as it freezes the water into ice.

#### Part 1: Heat removed to cool the water from 20°C to 0°C
The amount of heat removed to cool the water can be calculated using the specific heat formula:
\[
Q_1 = m \cdot c \cdot \Delta T
\]
Where:
- \(m = 1.71 \, \text{kg}\) (mass of water),
- \(c = 4186 \, \text{J/kg°C}\) (specific heat capacity of water),
- \(\Delta T = 20 - 0 = 20 \, \text{°C}\) (temperature change).

\[
Q_1 = 1.71 \cdot 4186 \cdot 20 = 143,160.8 \, \text{J}
\]

#### Part 2: Heat removed to freeze the water at 0°C
Next, the refrigerator removes the heat required to freeze the water. This can be calculated using the latent heat formula:
\[
Q_2 = m \cdot L_f
\]
Where:
- \(L_f = 334,000 \, \text{J/kg}\) (latent heat of fusion of water).

\[
Q_2 = 1.71 \cdot 334,000 = 571,140 \, \text{J}
\]

#### Total heat removed by the refrigerator:
\[
Q_{\text{total}} = Q_1 + Q_2 = 143,160.8 + 571,140 = 714,300.8 \, \text{J}
\]

### Step 2: Use the refrigerator's coefficient of performance (COP)
The refrigerator's coefficient of performance (COP) is the ratio of the heat removed (\(Q_{\text{total}}\)) to the work input. The work done by the refrigerator is:
\[
W_{\text{fridge}} = \frac{Q_{\text{total}}}{\text{COP}} = \frac{714,300.8}{2.13} = 335,774.1 \, \text{J}
\]

### Step 3: Compare with the space heater
The space heater produces \(2.68 \, \text{kW} = 2680 \, \text{J/s}\) of heat energy. To find the time it would take for the space heater to produce the same amount of heat as the refrigerator (in terms of work), we divide the energy by the heater's power:
\[
t = \frac{W_{\text{fridge}}}{\text{power of heater}} = \frac{335,774.1}{2680} \approx 125.3 \, \text{seconds}
\]

### Final Answer:
The space heater would need to run for approximately **125.3 seconds** (or about **2 minutes and 5 seconds**) to produce the same amount of heat as the refrigerator does when freezing the water.

Would you like further details or have any questions about the steps? Here are some related questions to expand the topic:

1. How is the coefficient of performance (COP) of a refrigerator defined?
2. What factors affect the COP of a refrigerator or heat pump?
3. How would this calculation change if the mass of water or temperature difference was different?
4. What is the difference between specific heat and latent heat?
5. How can energy efficiency be improved in both refrigerators and heaters?

**Tip:** When calculating energy changes involving phase transitions, always consider both the temperature change and the energy required for the phase change itself.

How long would a 2.68 kW space heater have to run to put into a kitchen the same amount of heat as a refrigerator (coeff of performance = 2.13) does when it freezes 1.71 kg of water at 20°C into ice at 0°C?

Discover how to calculate the time it takes for a 2.68 kW space heater to match the heat removed by a refrigerator (COP = 2.13) as it freezes 1.71 kg of water from 20°C to ice at 0°C. This solution involves thermodynamics concepts, including specific heat, latent heat, and energy efficiency. Suitable for Grades 10-12.

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