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