What is the heat change when a 26.8 g sample of water [C_water = 4.184 J/(g°C)] is cooled from 75.6°C to 22.1°C?

Question

Quizplus · Accepted Answer

Using the equation q = mCΔT, where q is the heat change, m is the mass, C is the specific heat capacity, and ΔT is the temperature change, we can calculate:

q = (26.8 g)(4.184 J/(g°C))(22.1°C - 75.6°C)
q = (26.8 g)(4.184 J/(g°C))(-53.5°C)
q = -6.00 x 10^3 J

Since the temperature is decreasing, the heat change is negative, so the answer is B: -6.00 x 10^13 J.

What Is the Heat Change When a 26