Question

The specific heat capacity of gold is 0.128 J/g ºC. How much heat would be needed to warm 250.0 grams of gold from 25.0 ºC to 100.0 ºC? A. 2.4 x10-3J B. -2.40 x 103J C. 2.40 x 103J D. -2.40 x 10-3J

Answer 1

C. 2.40 × 10³ J

Step-by-step explanation:

The formula for the amount of heat (q) absorbed is

q = mCΔT

1. Calculate ΔT

ΔT = 100.0 °C - 25.0 °C = 75.0 °C

2. Calculate q

q₂ = mCΔT = 250.0 g × 0.128 J·°C⁻¹g⁻¹ × 75.0 °C = 2.40 × 10³ J