Question

A hunk of up gold weighs 800 gram to heat the gold up to a temperature of 200 Kelvin if the gold absorbs 12 kilojoules of energy what was the initial temperature of the Gold? The specific heat capacity of gold is 0.129 grams times C

Answer 1

The initial temperature of the gold was 84 K.

Step 1. Calculate the temperature change.

The formula for the heat (q) absorbed is

q = mCΔT

We can solve this equation to get

ΔT = q/(mC)

q = 12 000 J; m = 800 g; C = 0.129 J·K⁻¹g⁻¹

∴ ΔT = 12 000 J/(800 g × 0.129 J·K⁻¹g⁻¹) = 116 K

Step 2. Calculate the initial temperature

ΔT = T_f – T_i = 200 K – T_i = 116 K

T_i = 200 K – 116 K = 84 K