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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