An 80 g sample of water is heated from 20°C to 80°C. Using c = 4.18 J/g°C, how many joules of heat are required?

An 80 g sample of water is heated from 20°C to 80°C. Using c = 4.18 J/g°C, how many joules of heat are required?

• A. 20,064 J
• B. 16,000 J
• C. 8,040 J
• D. 240 J

Answer: (A)

The main idea is that the heat needed to raise the temperature of a substance is found with Q = m c ΔT, where m is mass, c is the specific heat capacity, and ΔT is the temperature change. For water, c is 4.18 J/g°C. The mass is 80 g, and the temperature rises from 20°C to 80°C, so ΔT = 60°C. Plugging in: Q = 80 g × 4.18 J/g°C × 60°C. Compute 4.18 × 60 = 250.8, then 250.8 × 80 = 20,064 J. So the water requires 20,064 joules of heat. The other numbers would come from using a different ΔT or omitting a factor (mass or c), which isn’t consistent with the given data.

The main idea is that the heat needed to raise the temperature of a substance is found with Q = m c ΔT, where m is mass, c is the specific heat capacity, and ΔT is the temperature change. For water, c is 4.18 J/g°C. The mass is 80 g, and the temperature rises from 20°C to 80°C, so ΔT = 60°C. Plugging in: Q = 80 g × 4.18 J/g°C × 60°C. Compute 4.18 × 60 = 250.8, then 250.8 × 80 = 20,064 J. So the water requires 20,064 joules of heat. The other numbers would come from using a different ΔT or omitting a factor (mass or c), which isn’t consistent with the given data.