What amount of energy (heat) is required to warm a 65.0-g sample of liquid water from 25.0ºC to 96.5ºC? The specific heat capacity of water is 4.184 J/gºC.

Question

Quizplus · Accepted Answer

The energy required is calculated using the formula q = mcΔT, where m = 65.0 g, c = 4.184 J/gºC, and ΔT = 96.5ºC - 25.0ºC = 71.5ºC. Plugging in these values gives q = 65.0 g × 4.184 J/gºC × 71.5ºC = 1.94 × 10⁴ J.

What Amount of Energy (Heat) Is Required to Warm a 65.0-G