A 100 g water sample has specific heat 4.18 J/(g·°C). How much heat raises it by 5 °C? Assume constant specific heat and no phase change.

A) 2090 J

B) 4180 J

C) 1045 J

D) 418 J

Answer: A

Explanation: At approximately constant specific heat, q = mcΔT. Multiply 100 g by 4.18 J/(g·°C) and 5 °C to obtain 2090 J.