Specific heat capacity determines cooling rate

Steps:

• Equal masses of copper and water start at the same high temperature and cool by losing heat to surroundings.
• Heat loss rate depends on surface area, emissivity, and temperature difference, assumed similar for both.
• Temperature change ΔT = Q / (m c), where Q is heat lost, m is mass, c is specific heat capacity.
• For same m and Q, smaller c causes larger ΔT, so faster temperature drop.

Why B is correct:

• Specific heat capacity c is heat needed to raise 1 kg by 1°C; copper's c (385 J/kg°C) is smaller than water's (4186 J/kg°C), so copper cools faster per heat lost.

Why the others are wrong:

• A: Larger c would make copper cool slower, opposite of observed.
• C: Specific latent heat involves phase changes, irrelevant here as no phase change occurs during cooling.
• D: Smaller latent heat also irrelevant without phase change.

Final answer: B