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Answer :
Final answer:
To achieve a final temperature of 100°F in the tank, 13.4 gallons of water at 212°F must be added to 75 gallons of water at 80°F. This is based on the principle that the heat gained by the cooler water equals the heat lost by the hotter water, using the formula for heat transfer in thermodynamics.
So, the correct answer is B. 13.4 gal.
Explanation:
The question involves calculating the amount of hot water needed to raise the temperature of a cooler water sample to a specific temperature, a common problem in thermodynamics and heat transfer. To find the solution, we can use the equation Q = mcΔT, where Q is the heat energy transferred, m is the mass or volume of water in this case, c is the specific heat capacity (approximately 4.184 J/g°C or 1 Btu/lb°F for water), and ΔT is the change in temperature. However, in this case, because we are mixing liquids at different temperatures, we can set the heat gained by the cooler water equal to the heat lost by the hotter water without having to calculate Q explicitly:
Let V be the volume of water at 212°F to be added to the 75 gallons of 80°F water to achieve a final temperature of 100°F.
(75 gal)(100°F - 80°F) = (V)(212°F - 100°F)
Now solve for V:
(75 gal)(20°F) = (V)(112°F)
1500 gal°F = V × 112°F
V = 1500 gal°F / 112°F
V = 13.4 gallons
Therefore, the answer is 13.4 gallons of water at 212°F must be added to 75 gallons of water at 80°F to produce water at 100°F in a tank, which is option B.
So, the correct answer is B. 13.4 gal.
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