How much heat is required to increase the temperature of 20 lb of water from 32°F to 212°F?
A) (20 × 144 ) + (20 × 180 ) + (20 × 970 ) = 25,880 BTU B) 212°F - 32°F = 180 BTU
C) 20 lb × 180°F = 3,600 BTU D) 20 lb × 0.5 × 180°F = 1,800 BTU
C
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