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WAEC Physics 2019 Paper

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Question : 17

Total: 49

Calculate the quantity of heat needed to change the temperature of 60g of ice at 0 °C to 80 °C. [specific latent heat of fusion of ice= 3.36 x 10

^5

^{-1}

specific heat capacity of water 4.2 x 10

^3

^{-1}

^{-1}

]

4.80 kJ
20.16 kJ
40.32 kJ
22.17 kJ

Solution:

Concept:

Converting ice to water requires latent heat, and then heating that water requires sensible heat. The total heat is the sum of both.

Explanation:

Mass of ice,

m = 60g = 0.06kg

.

Stage 1: Melt ice at

0^\circ C

to water at

0^\circ C

.

Heat needed,

Q_1 = mL_f = 0.06 \times 3.36 \times 10^5

.

Q_1 = 2.016 \times 10^4J = 20.16kJ

.

Stage 2: Raise water temperature from

0^\circ C

to

80^\circ C

.

Heat needed,

Q_2 = mc\Delta T = 0.06 \times 4.2 \times 10^3 \times (80 - 0)

.

Q_2 = 0.06 \times 4.2 \times 10^3 \times 80 = 2.016 \times 10^4J = 20.16kJ

.

Total heat,

Q = Q_1 + Q_2 = 20.16 + 20.16 = 40.32kJ

.

Answer:

40.32kJ

, which is option C.

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