Answer:
The answers to the question are
a. 166.64 ° F
b. 217990.08 J/hour or 60.55 J/s = 60.55 watts
c. 13.C
Explanation:
a. To solve the question we list out the given variables thus
mass of grain = 16.5 lbs
Temperature of grain = 67 °F
Volume of hot water = 5 gals = 0.02273 m³
Equilibrium temperature of the mixture = 154 °F
Specific heat capacity of the grain = 0.44 times specific heat capacity of water
Therefore we have
Heat supplied by hot water = heat gained by mixture
Density of the water = 997 kg/m³ which gives
Therefore the mass of the water = (Density of the water) × (Volume of the water) = (997 kg/m³) × (0.02273 m³) = 22.66181 kg
Therefore the heat supplied by the water =22.66 kg×1000 g/kg ×4.2 J/g°C×(Tₓ -67.78 °C) = 7.48 kg×1000 g/kg×0.44×4.2 J/g°C×(67.78 -19.44)
= 95172 × (Tₓ -67.78 °C) =668205.7536 J
(Tₓ -67.78 °C) = 7.02 from where Tₓ = 74.80 °C = 166.64 ° F
The initial temperature (strike temperature) of the hot water = 74.80 °C = 166.64 ° F
b. Where the mixture lost two degrees we have
22.66 kg×1000 g/kg ×4.2 J/g°C×2 °C + 7.48 kg×1000 g/kg×0.44×4.2 J/g°C×2 °C = 217990.08 J therefore the average energy lost per unit time = 217990.08 J/hour or 60.55 J/s
c. To find out how much it cost we have
Heat energy required to raise 5 gallons of water from 110 °F to 166.64 °F we have
22.66 kg×1000 g/kg ×4.2 J/g°C×(74.8 °C-43.33 °C) = 2994745.92 J
Energy lost during the heating = 10% = 299474.59 J
Total energy supplied 2994745.92 J + 299474.59 J = 3294220.5 J
Time for heating = 47 minutes, therefore rate of energy consumption = (3294220.5 J)/ (47×60) = 1168.163 Watt 1.168 kW
Cost of energy = 15.C per kilowatt-hour therefore 1.168 kW for 47 minutes will cost
1.168 kW ×47/60×15 = 13.C
therefore it cost 13.C to heat the 5 gallons of tap water initially at 110 ° F to the strike temperature 166.64 °F
