The specific latent heat of fusion of ice is 336 kJ/kg. The heat required to melt 2 kg of ice at 0°C is: MCQ with Answer and Explanation

The specific latent heat of fusion of ice is 336 kJ/kg. The heat required to melt 2 kg of ice at 0°C is:
A. 168 kJ
B. 1344 kJ
C. 336 kJ
D. 672 kJ
Answer: Option D
Solution (By JKSSB Mock Tests)
Heat for phase change Q = m·L_f, where L_f is latent heat of fusion. Thus Q = 2 kg × 336 kJ/kg = 672 kJ. This direct application tests calorimetry fundamentals. Memory aid: 'Latent heat: Q = mL; no temperature change during phase transition'. Competitive exams frequently test such calculations with standard values. Always ensure units match (kg and kJ/kg here); convert if necessary. This problem assesses basic formula application skills essential for thermodynamics sections.

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Question #1
A cell of emf 2 V and internal resistance 0.5 Ω is connected to a 3.5 Ω resistor. Terminal voltage is
A. 2 V
B. 0.25 V
C. 1.75 V
D. 1.5 V

Correct Answer: Option C


Explanation:
I = E/(R+r) = 2/(3.5+0.5) = 2/4 = 0.5 A. Terminal V = IR = 0.5×3.5 = 1.75 V. Or E - Ir = 2 - 0.5×0.5 = 1.75 V.

This question belongs to: Science Physics
Question #2
J. J. Thomson is credited with the discovery of:
A. Proton
B. Electron
C. Neutron
D. Nucleus

Correct Answer: Option B


Explanation:
J. J. Thomson discovered the electron in 1897 through cathode ray tube experiments, measuring charge-to-mass ratio (e/m) of cathode rays. Proton: Rutherford; neutron: Chadwick; nucleus: Rutherford's gold foil experiment. Memory aid: 'Thomson = plum pudding model (electrons in positive sphere)'. This historical question tests knowledge of subatomic particle discoveries, frequently examined in competitive exams. Always recall the experimental methods: Thomson (cathode rays), Millikan (oil drop for e), Rutherford (alpha scattering).

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Question #3
An echo returned in 0.5 s. If speed of sound is 340 m/s, the distance of reflecting surface is
A. 170 m
B. 85 m
C. 680 m
D. 340 m

Correct Answer: Option B


Explanation:
d = (v×t)/2 = (340×0.5)/2 = 170/2 = 85 m. Minimum distance for echo perception is about 17 m. Remember divide by 2 for one-way distance.

This question belongs to: Science Physics