Test Index

JAMB Physics 2012 Paper

Question : 17

Total: 42

Calculate the temperature change when 500J of heat is supplied to 100g of water

Solution:

Concept:

Heat gained by a substance is related to its mass, specific heat capacity, and temperature change by

Q = mc\Delta\theta

Explanation:

Convert mass to kilograms:

m = 100\text{ g} = 0.1\text{ kg}

Use specific heat capacity of water:

c = 4200\text{ J kg}^{-1}\text{ K}^{-1}

Make temperature change the subject:

\Delta\theta = \frac{Q}{mc}

Substitute values:

\Delta\theta = \frac{500}{0.1 \times 4200} = \frac{500}{420}

Calculate:

\Delta\theta \approx 1.19^{\circ}\text{C} \approx 1.2^{\circ}\text{C}

Answer:

1.2^{\circ}\text{C}

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