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What is the total amount of heat absorbed by 100.0 grams of water when the temperature of the water is increased from 30.0°C to 45.0°C?
(1) 418 J (3) 12 500 J
(2) 6270 J (4) 18 800 J
The total amount of heat is the same as the energy. Because no phase change is occurring, use the formula Q = mc(delta)T. Q is the energy, m is the mass, c is the specific heat capacity of water as a liquid, and (delta)T is the change in temperature. So: Q = (100.0 g)(4.186 J/g°C)(15°C); Q = 6279 Joules.
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