Question:
Given the rate of thermal energy transfer is 2.7kW, the volume of the water tank is 4.5m^3, the water is at a temperature of 28ºC, density of water is 1000kgm-3 & c=4200Jkg-1K-1. Calculate the rise in water temperature that the heater could produce in 1hr.
Answer:
The equations needed to solve this problem are outlined below. E = mc(T2 – T1) d = m/v P = E/tYou first need to work out the mass of the water. m = 1000 x 4.5 = 4500kg You then need to work out the amount of energy produced by the heater in one hour. 2700 = E / (60 x 60)E = 9.72 x 106JThe change in temperature of the water can then be calculated.9.72 x 106 = 4500 x 4200 x (T2 – T1) Change in T = 0.51K
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