Question

A 200-L electric water heater uses 2.0 kW. Assuming no heat loss, how many hours would it take to heat the water in this tank from 23°C to 75°C? The specific heat of water is 4186 J/kg • K and its density is 1000 kg/m³.

answers:

5.0 hours

8.0 hours

6.0 hours

7.0 hours

Answer 1

Because the density of water is approximately 1 g/L, we are tryingto heat up about 2 kg of water. Thus, using the formula

q = mcΔt,

we want to find the amount of energy required to complete theprocess.

q = (2 kg)(4186 J/kg-K)(75-23 K) = 4.353 * 10^3 kJ ofheat.

Now your heater provides heat at a rate of 2 kW which equals 2kJ/second. Thus the time we require to heat up the water is givenby:

(4.353* 10^3 kJ)/(2 kJ/sec) = 2.176 * 10^5 seconds.

(2.176 * 10^5 seconds)/(3600 seconds/hour) = 6.0444 hours,which matches your answer of 6.0 hours.

Then 6.0 hours in a correct answer

I hope help you !!