KCSE 2025 Physics P1 Q15 — Thermal Physics (specific heat, latent heat and mixtures)

KCSE 2025 Form 4 Thermal Physics

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The Question

“This heat question comes in several parts. (a) Define specific heat capacity. (b)(i) Explain what happens, in terms of heat flow, when an ice cube is placed in warm water. (b)(ii) State two ways in which the melting point of ice can be lowered. (c) A block of ice at 0 °C is supplied with 2000 J of heat. Taking the specific latent heat of fusion of ice as L_f = 3.36×10⁵ J/kg, determine the mass of ice that melts. (d) 25 g of water at 50 °C is mixed with 75 g of water at 20 °C in a container of negligible heat capacity. Determine the final temperature of the mixture.”

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(a) Define specific heat capacity

Specific heat capacity is the quantity of heat needed to raise the temperature of a unit mass of a substance by one degree — that is, by one kelvin (equivalently one degree Celsius). It measures how much heat a material must absorb per kilogram for each degree of temperature rise.

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(b)(i) Heat flow when ice is placed in water

When an ice cube is placed in the warmer water, heat flows from the warmer water to the colder ice because heat always travels from a hotter body to a colder one. This heat supplies the latent heat of fusion of the ice, which breaks the bonds holding the solid together, and so the ice melts. Notice the temperature of the ice does not rise while it melts — the heat goes into changing its state, not its temperature.

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(b)(ii) Two ways to lower the melting point of ice

The melting point of ice can be reduced in two ways: first, by increasing the pressure on the ice; and second, by adding impurities such as salt to it. Both are why gritting roads with salt and pressing on ice can make it melt below 0 °C.

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(c) Set up the latent-heat equation

When ice melts at a steady 0 °C, the heat supplied does not change its temperature — it is all latent heat of fusion. The heat is Q = m·L_f, where m is the mass melted and L_f is the specific latent heat of fusion. Rearrange to make the mass the subject.

Q=mLf ⇒ m=QLfQ = m L_f \ \Rightarrow \ m = \frac{Q}{L_f}
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(c) Substitute the values and solve

Put in the heat supplied, 2000 J, and the latent heat of fusion of ice, 3.36×10⁵ J/kg. Dividing gives the mass of ice that melts.

m=20003.36×105m = \frac{2000}{3.36 \times 10^5}
5.95×10−3 kg5.95 \times 10^{-3}\ \mathrm{kg}
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(d) Apply the principle of mixtures

When 25 g of hot water at 50 °C is mixed with 75 g of cold water at 20 °C, the heat lost by the hot water equals the heat gained by the cold water. Both are water, so the specific heat capacity is the same on each side and cancels out, leaving only the masses and temperature changes. Writing T for the common final temperature gives the balance below.

25 g at 50∘C + 75 g at 20∘C25\,\text{g at }50^{\circ}\text{C} \ + \ 75\,\text{g at }20^{\circ}\text{C}
25(50−T)=75(T−20)25(50 - T) = 75(T - 20)
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(d) Solve for the final temperature

Divide both sides by 25 to simplify, then expand the brackets and collect the terms in T. This leaves 110 = 4T, so dividing by 4 gives the final temperature of the mixture.

50−T=3(T−20)50 - T = 3(T - 20)
110=4T110 = 4T
T=27.5 ∘CT = 27.5\,^{\circ}\text{C}

Final Result

(a) Specific heat capacity is the heat needed to raise the temperature of unit mass of a substance by one kelvin. (b)(i) Heat flows from the warm water into the colder ice, supplying its latent heat of fusion so it melts. (b)(ii) Increase the pressure on the ice, or add impurities such as salt. (c) The mass of ice melted is 5.95×10⁻³ kg. (d) The final temperature of the mixture is 27.5 °C.

Why this method works

Two distinct ideas carry this question. First, heat added to a substance either raises its temperature (governed by specific heat capacity, Q = mcΔθ) or changes its state at constant temperature (governed by latent heat, Q = mL). While ice melts, its temperature is stuck at 0 °C, so all the supplied energy is latent heat of fusion — that is why part (c) uses Q = mL_f and not Q = mcΔθ. Second, the method of mixtures is really the conservation of energy: in an insulated container no heat escapes, so every joule the hot water loses is a joule the cold water gains. Because both liquids are water with the same specific heat capacity, that quantity cancels and the balance reduces to masses times temperature changes. The final temperature settling nearer 20 °C than 50 °C is exactly what you should expect, because there is three times as much cold water as hot water pulling the mixture toward the lower temperature.

✓

For (d), check energy balance: hot water loses 25(50−27.5) = 562.5 units of heat and cold water gains 75(27.5−20) = 562.5 units — they match, confirming T = 27.5 °C.