A piece of copper of mass 20g at a temperature of 110°C was dropped into a mixture of ice and water at 0°C. If…

Analyze the question: Hot copper loses heat as it cools from 110°C to 0°C. That heat melts some of the ice, and the mixture stays at 0°C throughout.

Concept:

• Heat lost by copper = m c Δθ
• Heat gained by ice = m_ice × L
• At 0°C, heat lost by copper = heat used to melt ice

Solution:

Heat lost by copper: Q = m c Δθ = 20 × 0.4 × (110 − 0) = 880 J

Heat gained by melting ice: Q = m_ice × 330

Equate the two: m_ice × 330 = 880 m_ice = 880 / 330 = 2.67 g ≈ 2.7 g

Answer: The mass of ice that melted is about 2.7 g, which is option D.

Tip: When the final temperature is the melting point (0°C), no temperature change happens to the ice-water mixture. All the heat from the hot object goes into melting ice, so just use Q = mcΔθ = mL. Also, 880/330 = 8/3 ≈ 2.67, so you can round it to 2.7 quickly.