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JAMB Physics 2021 Paper

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Question : 32

Total: 40

Calculate the mass of ice that would melt when 2kg of copper is quickly transferred from boiling water to a block of ice without heat loss (specific heat capacity of copper = 400Jkg

^{1}

^{-1}

, latent heat of fusion of ice = 3.3 x 10

^{5}

^{-1}

)

$\frac{8}{33}$ kg
$\frac{3.3}{80}$ kg
$\frac{80}{33}$ kg
$\frac{3.3}{8}$ kg

Solution:

Concept:

Heat lost by hot copper is completely used to melt ice without any loss to the surroundings.

Explanation:

Copper is taken from boiling water, so its initial temperature is

100^\circ\text{C}

.

The ice is at

0^\circ\text{C}

, therefore the temperature drop of copper is

\Delta\theta = 100 - 0 = 100\,\text{K}

.

Heat lost by copper:

Q_c = m_c c_c \Delta\theta

Substitute

m_c = 2\,\text{kg}

,

c_c = 400\,\text{Jkg}^{-1}\text{K}^{-1}

:

Q_c = 2 \times 400 \times 100 = 80000\,\text{J}

Let the mass of ice melted be

m_i

.

Heat gained by ice:

Q_i = m_i L_f

, with

L_f = 3.3 \times 10^5\,\text{Jkg}^{-1}

Equating heat lost and heat gained:

m_i \times 3.3 \times 10^5 = 80000

m_i = \frac{80000}{3.3 \times 10^5} = \frac{8}{33}\,\text{kg}

Answer:

The correct option is A.

\frac{8}{33}\,\text{kg}

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