A 0.020\text{ kg} sample of a liquid metal at its melting point of 500^\circ\text{C} solid… — JAMB UTME

A $0.020\text{ kg}$ sample of a liquid metal at its melting point of $500^\circ\text{C}$ solidifies completely as thermal energy is extracted from it at a constant rate of $10\text{ W}$ . If the specific latent heat of fusion of the metal is $2.0 \times 10^4\text{ J kg}^{-1}$ , how long does the solidification process take?

$40\text{ s}$ Answer
B
$400\text{ s}$
C
$2000\text{ s}$
D
$20000\text{ s}$

Answer

The solidification process takes

40\text{ s}

The thermal energy released when a substance solidifies at its melting point is given by

Q = m L_f

. Substituting

m = 0.020\text{ kg}

and

L_f = 2.0 \times 10^4\text{ J kg}^{-1}

gives

Q = 400\text{ J}

. Dividing this energy by the constant rate of heat removal (

10\text{ W}

) yields

t = \frac{400\text{ J}}{10\text{ W}} = 40\text{ s}

Step-by-Step Solution

Calculate the total thermal energy (

Q

) released during phase change at constant temperature

Q = m L_f = 0.020\text{ kg} \times 2.0 \times 10^4\text{ J kg}^{-1} = 400\text{ J}

Phase change occurs at a constant temperature, so thermal energy depends solely on mass and specific latent heat of fusion.

Calculate time (

t

) required using the power rate (

P

)

t = \frac{Q}{P} = \frac{400\text{ J}}{10\text{ W}} = 40\text{ s}

Power is defined as energy transferred per unit time (

P = Q / t

Key Concept

Latent Heat of Fusion and Energy Balance

Question

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

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