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JAMB Physics 2023 Paper

Question : 21

Total: 65

-10^oC

10^oC

10^5

Jkg^-1

Jkg^-1

k^-1

Solution:

Quantity of heat required to raise the temperature of the ice cube from $-10^oC$ to $0^oC$

⇒H=mc►ø = 0.02 × 2100 × 0 -(-10)

⇒ H = 0.02 × 2100 × (0 + 10) = 0.02 × 2100 × 10

⇒ H = 420 J

quantity of heat required to melt ice at

0^oC

⇒ H = mL = 0.02 × 3.34 ×

10^5

⇒H=6680 j

Quantity of heat required to raise the temperature of the melted ice cube (water) from $0^oC$ to $-10^oC$

⇒H=mc►ø = 0.02 × 4200 × (10-0)

⇒H = 0.02 × 4200 × 10

⇒H = 840j

;420 + 6680 + 840 = 7940 j

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