Answer:
The energy in kJ is 8558.16 kJ.
Explanation:
Data presented in the problem:
Water is heated from 70 (T1) to 200 °F (T2).
Volume (V) of the water is 1 ft3.
It is required for the specific heat of water(HW), which is 1 BTU/lb°F.
First, we need to calculate the mass of water (M) presented in the process. Water density (D) is 62. 4 lb/ft3.
M = V*D = (1ft3)*(62.4 lb/ft3) = 62.4 lb Water.
.After that, we can calculate the heat required (Q).
Q = M*HW*(T2 - T1) = (62.4 lb)*(1 BTU/lb°F)(200 °F - 70°F)
Q = 62.4 * 130 BTU = 8112 BTU.
Q is converted to kJ units using the conversion factor 1 BTU = 1.055 kJ
Q = 8112 BTU * (1.055 kJ/1BTU) = 8558.16 kJ.
Finally, the energy required is 8558.16 kJ.
