If 418 J of heat is added to 100 g of water with a specific heat capacity of 4.18 J/g°C, what is the resulting temperature rise?

• A. 1.0°C
• B. 0.1°C
• C. 10.0°C
• D. 0.418°C

Answer: (A)

Heat added to a substance relates to how much its temperature changes through q = m c ΔT. Here, q = 418 J, mass m = 100 g, and the specific heat c for water is 4.18 J/g°C.

Calculate the product m c: 100 g × 4.18 J/g°C = 418 J/°C. Then solve for the temperature rise: ΔT = q/(m c) = 418 J ÷ 418 J/°C = 1 °C.

So the water temperature increases by 1.0°C. Quick checks: a 0.1°C rise would need about 41.8 J, a 10°C rise would need about 4180 J, etc., which don’t match the given heat.