What is the Possible Energy Collected in kJ if the power striking an oven 53.2 cm by 44.5 cm is 1071.5 watt/meters squared over 60.9 minutes?

Question

Please round to 3 sig figs but use the following format. 

Ex: XXX0 if greater than 1000 or XXX if less than 1000. 

X.XXEX also works either way

Answer 1

So we are given the dimensions/area of the oven (53.2 cm by 44.5 cm), the power per square meter (1071.5 watts/square meter), and the time the power is applied (60.9 minutes).

First, we can determine the total power applied by multiplying the area of the oven by the power per square meter. We need to convert the dimension into meters rather than centimeters: 0.532 meters by 0.445 meters. Multiplying these values together gives an area of about 0.237 square meters. Now we determine the total power applied: (0.237 square meters * 1071.5 watts/square meter)= about 254 watts.

Now, we must identify that 1 watt is equal to 1 Joule per second (J/s), so 254 watts = 254 J/s. Now we convert 60.9 minutes into seconds: about 3650 seconds. Therefore we can find the total Joules collected by multiplying: (254 J/s * 3650 s)= 927100 J = about 927 kJ.