Question

A room contains 75 kg of air at 100 kPa and \(15^{\circ} \mathrm{C}\) The room has a 250 -W refrigerator (the refrigerator consumes \(250 \mathrm{W} \text { of electricity when running }),\) a \(120-\mathrm{W} \mathrm{TV},\) a 1.8-kW electric resistance heater, and a 50-W fan. During a cold winter day, it is observed that the refrigerator, the TV, the fan, and the electric resistance heater are running continuously but the air temperature in the room remains constant. The rate of heat loss from the room that day is \((a) 5832 \mathrm{kJ} / \mathrm{h}\) (b) \(6192 \mathrm{kJ} / \mathrm{h}\) \((c) 7560 \mathrm{kJ} / \mathrm{h}\) \((d) 7632 \mathrm{kJ} / \mathrm{h}\) \((e) 7992 \mathrm{kJ} / \mathrm{h}\)

Short answer

Answer: The rate of heat loss from the room is 7992 kJ/h.

Step-by-step solution

Calculate the total power consumed by all appliances

First, we need to find the total power consumed by all the appliances combined. Add the power of the refrigerator, the TV, the fan, and the electric resistance heater together. Total power = 250 W (refrigerator) + 120 W (TV) + 50 W (fan) + 1800 W (heater) = 2220 W

Convert the total power to kilowatts

To have a clear picture of the total heat generated in the room, convert the total power from watts to kilowatts. Total power in kilowatts = 2220 W * (1 kW / 1000 W) = 2.22 kW

Calculate the total heat generated in a hour

Now, we need to find out the total heat generated per hour by the appliances in the room. To do that, multiply the total power in kW with the time in hours. Total heat generated per hour = Total power in kilowatts * Time (hours) Total heat generated per hour = 2.22 kW * 1 hour = 2.22 kWh

Convert the heat into kJ/h

Now, we need to convert the total heat generated in a hour from kWh to kJ/h. We will use the conversion that 1 kWh = 3600 kJ. Total heat generated per hour in kJ/h = 2.22 kWh * (3600 kJ/ 1 kWh) = 7992 kJ/h So, the rate of heat loss from the room that day is \((e) 7992 \mathrm{kJ} / \mathrm{h}\).