# SPM Physics Heat - How do we do this?

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1. Some-ice cubes are added to 400 \mathrm{~g} of water at 30^{\circ} \mathrm{C}. The minimum temperature of the mixture of ice and water when all the ice cubes have melted is 8^{\circ} \mathrm{C}. What is the mass of the ice cubes used? [Specific heat capacity of water =4200 \mathrm{~J} \mathrm{~kg}^{-1}{ }^{\circ} \mathrm{C}^{-1}, Specific latent heat of fusion of ice \left.=3.34 \times 10^{5} \mathrm{~J} \mathrm{~kg}^{-1}\right] Beberapa kiub ais ditambah ke dalam 400 \mathrm{~g} air pada suhu 30^{\circ} \mathrm{C}. Suhu minimum campuran ais dan air apabila semua kiub ais telah melebur ialah 8^{\circ} \mathrm{C}. Berapakah jisim kiub ais yang digunakan? [Muatan haba tentu air =4200 \mathrm{~J} \mathrm{~kg}^{-1}{ }^{\circ} \mathrm{C}^{-1}, Haba pendam tentu ais \left.=3.34 \times 10^{5} \mathrm{~J} \mathrm{~kg}^{-1}\right]
A 101 \mathrm{~g}
C 255 \mathrm{~g}
B 111 \mathrm{~g}
D 402 \mathrm{~g}

First we perform an energy balance for both the ice and water. In this case, energy lost by the water (at higher temperature) must be EQUAL to the energy gained by the ice (at lower temperature, assuming 0

1. Energy gained by ice = Energy absorbed for latent heat of fusion (ice → water) + Sensible heat absorbed to raise temperature of water from 0 °C to 8 °C
1. Energy lost by water = Sensible heat lost to lower temperature from 30 °C to 8 °C.

First we will do the water side (400 g = 0.4 kg)

Energy lost by water, Q = mcpΔT
Q = 0.4(4200)(30-8) = 36,960 J

Then, we do the ice side (mass = unknown, let it = m). Since energy lost by water must be absorbed by ice, Qi = Q = 36,960 J.

Energy gained by ice, Qi = m x latent heat + mcpΔT = m(latent heat + cpΔT)
Sensible heat portion involve heating the cold water from 0 to 8 °C after melting
Q = 36,960 J = m( 3.34 x 10^5 + 4200(8-0) )
m = 0.1005 kg = 100.5 g ~= 101 g

Therefore, answer is A. 101 g

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