Answer:
(a) The heat transfer rate is 882 watts.
(b) The output temperature of the water is 44 ºC.
Explanation:
Statement is incomplete. Complete description shall be described below:
A solar water panel heats the water flowing through from 14 to 35° C when water flows through at the rate of 0.010 kilograms per second.
(a) Calculate the energy per seconds transferred by the hot water from the solar panel.
(b) Estimate the output temperature of the hot water if the flow is reduced to 0.007 kilograms per second. The specific heat of water is 4200 joules per kilogram-degree Celsius.
(a) The heat transfer rate received by the solar water panel equals the heat transfer rate associated with the change in sensible heat of the water, both measured in kilowatts:
[tex]\dot Q = \dot m \cdot c\cdot (T_{o}-T_{i})[/tex] (1)
Where:
[tex]\dot Q[/tex] - Heat transfer rate, measured in watts.
[tex]\dot m[/tex] - Mass flow, measured in kilograms per second.
[tex]T_{i}[/tex], [tex]T_{o}[/tex] - Inlet and outlet temperatures of water, measured in degrees Celsius.
If we know that [tex]\dot m = 0.010\,\frac{kg}{s}[/tex], [tex]c = 4200\,\frac{J}{kg\cdot ^{\circ}C}[/tex], [tex]T_{i} = 14\,^{\circ}C[/tex] and [tex]T_{o} = 35\,^{\circ}C[/tex], then the heat transfer rate is:
[tex]\dot Q = \left(0.010\,\frac{kg}{s} \right)\cdot \left(4200\,\frac{J}{kg\cdot ^{\circ}C} \right)\cdot (35\,^{\circ}C-14\,^{\circ}C)[/tex]
[tex]\dot Q = 882\,W[/tex]
The heat transfer rate is 882 watts.
(b) Let suppose that heat transfer rate is same of exercise (a). If we know that [tex]\dot Q = 882\,W[/tex], [tex]\dot m = 0.007\,\frac{kg}{s}[/tex], [tex]c = 4200\,\frac{kJ}{kg\cdot ^{\circ}C}[/tex] and [tex]T_{i} = 14\,^{\circ}C[/tex], the output temperature of the water is:
[tex]T_{o} = T_{i}+\frac{\dot Q}{\dot m\cdot c}[/tex]
[tex]T_{o} = 14\,^{\circ}C + \frac{882\,W}{\left(0.007\,\frac{kg}{s} \right)\cdot \left(4200\,\frac{J}{kg\cdot ^{\circ}C} \right)}[/tex]
[tex]T_{o} = 44\,^{\circ}C[/tex]
The output temperature of the water is 44 ºC.
