Question

A room is heated by a baseboard resistance heater. When the heat losses from the room on a winter day amount to \(7000 \mathrm{~kJ} / \mathrm{h}\), it is observed that the air temperature in the room remains constant even though the heater operates continuously. Determine the power rating of the heater in \(\mathrm{kW}\).

Short answer

Answer: The power rating required for the baseboard resistance heater is approximately 1.944 kW.

Step-by-step solution

Identify the given information

The heat loss from the room is given in kJ/h, which is 7000 kJ/h.

Convert the heat loss from kJ/h to kW

In order to find the power rating, we need to convert the value from kJ/h to kW. To do this, we will use the following conversion factors: 1 kW = 1 kJ/s 1 hour = 3600 seconds So, we have: $$ Heat Loss (kW)= \frac{Heat Loss (kJ/h)}{3600 (s/h)} $$ Now, substitute the given value: $$ Heat Loss (kW)= \frac{7000\:kJ/h}{3600\:s/h} $$

Calculate the power rating of the heater

After substituting the values, we can now calculate the power rating required of the heater: $$ Heat Loss (kW)=\frac{7000\:kJ/h}{3600\:s/h}=1.944\:kW $$

Conclusion

The power rating required for the baseboard resistance heater to maintain a constant room temperature is approximately 1.944 kW.