To find out how much thermal energy is removed from the refrigerator during one hour of operation, we need to calculate the rate of heat transfer due to convection. The formula for convective heat transfer is given by:

$Q = h \cdot A \cdot (T_s - T_\infty)$

Where:

• $Q$ is the rate of heat transfer (W or J/s),
• $h$ is the convective heat transfer coefficient (18 W/m²K),
• $A$ is the surface area (0.5 m²),
• $T_s$ is the surface temperature of the heat exchanger (33°C),
• $T_\infty$ is the room temperature (20°C).

Step 1: Convert temperatures to Kelvin (optional since we are working with temperature differences, Celsius works too)

$T_s = 33°C, \quad T_\infty = 20°C$

Step 2: Calculate the temperature difference

$\Delta T = T_s - T_\infty = 33°C - 20°C = 13°C$

Step 3: Plug in the values into the heat transfer equation

$Q = 18 \, \text{W/m}^2\text{K} \cdot 0.5 \, \text{m}^2 \cdot 13 \, \text{K} = 117 \, \text{W}$

This means the refrigerator removes 117 watts of thermal energy to the room.

Step 4: Calculate the total energy removed during 1 hour of operation

Since 1 hour equals 3600 seconds, the total energy removed is: $Q_{\text{total}} = 117 \, \text{W} \cdot 3600 \, \text{s} = 421,200 \, \text{J} = 421.2 \, \text{kJ}$

Thus, the refrigerator removes 421.2 kJ of thermal energy during 1 hour of operation.

Would you like more details or have any questions?

5 Related Questions:

1. How does changing the surface area of the heat exchanger affect the amount of heat removed?
2. What would happen if the heat transfer coefficient increased?
3. How would the result change if the room temperature was higher?
4. Can you calculate the energy removed if the refrigerator runs for 2 hours?
5. How does convection differ from conduction in terms of heat transfer?

Tip:

When dealing with heat transfer problems, always double-check the units and make sure you're using the correct heat transfer equation for the type of transfer (conduction, convection, or radiation).