Specific Heat & Heat Capacity — NEET Physics MCQs with Solutions
Free NEET Physics Specific Heat & Heat Capacity MCQs with step-by-step solutions (20 questions). Part of Calorimetry & Thermal Properties. Practise online on Prepizo — no login needed.
▶ Practise Specific Heat & Heat Capacity online (free)
Questions with solutions
Q1 — Specific Heat & Heat Capacity · easy · theory
The specific heat capacity of a substance is the heat required to raise the temperature of:
A. One mole of the substance by 1°C
B. Unit mass of the substance by 1°C (or 1 K) ✓ Correct
C. The whole body by 1°C
D. Unit volume by 1°C
Solution: c = Q/(mΔT): heat per unit mass per unit temperature rise. Units: cal/g°C or J/kg·K.
Q2 — Specific Heat & Heat Capacity · easy · theory
The heat capacity (thermal capacity) C of a body is related to its specific heat c by:
A. C = mc²
B. C = m/c
C. C = c/m
D. C = mc ✓ Correct
Solution: Heat capacity = heat to raise the WHOLE body by 1°C: C = mc (units cal/°C or J/K).
Q3 — Specific Heat & Heat Capacity · easy · theory
The molar heat capacity C_m of a substance of molar mass M and specific heat c is:
A. C_m = M/c
B. C_m = c
C. C_m = c/M
D. C_m = M·c ✓ Correct
Solution: Molar heat capacity = heat per mole per degree = (mass of one mole) × (specific heat) = Mc.
Q4 — Specific Heat & Heat Capacity · easy · theory
The SI unit of specific heat capacity is:
A. J·kg/K
B. cal/g°C
C. J/kg·K ✓ Correct
D. J/K
Solution: SI: joule per kilogram per kelvin (J/kg·K). cal/g°C is the practical (CGS-style) unit; 1 cal/g°C = 4200 J/kg·K.
Q5 — Specific Heat & Heat Capacity · easy · theory
Among common substances, water has:
A. Zero specific heat
B. One of the highest specific heat capacities (1 cal/g°C) ✓ Correct
C. The lowest specific heat
D. A specific heat equal to that of ice
Solution: Water's exceptionally high specific heat (1 cal/g°C = 4200 J/kg·K) is why it is used as a coolant and moderates climate. (Ice is about half: 0.5 cal/g°C.)
Q6 — Specific Heat & Heat Capacity · medium · theory
The mechanical equivalent of heat J relates work W and heat Q by:
A. W = Q always
B. Q = JW, with J ≈ 4.186 cal/J
C. W = Q/J², J in joules
D. W = JQ, with J ≈ 4.186 J/cal ✓ Correct
Solution: J = W/Q ≈ 4.186 joule per calorie — the work needed to produce one calorie of heat (Joule's experiment).
Q7 — Specific Heat & Heat Capacity · medium · theory
During the free fall of a waterfall, if all the mechanical energy converts to heat, the temperature rise of the water depends on:
A. The mass of the water
B. The density of water only
C. Only the height of the fall (ΔT = gh/c) ✓ Correct
D. The width of the waterfall
Solution: mgh = mcΔT ⇒ ΔT = gh/c — independent of mass. Speed-trick: for h in metres, ΔT ≈ h/420 °C (using c = 4200 J/kg·K, g = 10).
Q8 — Specific Heat & Heat Capacity · medium · theory
A body's specific heat, in general:
A. Depends on the shape of the body
B. Is strictly constant at all temperatures
C. Is always zero at 0°C
D. Varies with temperature, though it is often taken constant over small ranges ✓ Correct
Solution: Specific heat is temperature-dependent (famously for water it varies slightly, minimum near 35°C), but is treated as constant for typical calorimetry ranges.
Q9 — Specific Heat & Heat Capacity · easy · numerical
The heat required to raise the temperature of 500 g of water from 20°C to 60°C is:
A. 20000 cal ✓ Correct
B. 10000 cal
C. 30000 cal
D. 40000 cal
Solution: Q = mcΔT = 500 × 1 × (60 − 20) = 20000 cal = 20 kcal.
Q10 — Specific Heat & Heat Capacity · easy · numerical
The heat needed to raise the temperature of 2 kg of water by 5 K (c_w = 4200 J/kg·K) is:
A. 4200 J
B. 8400 J
C. 21000 J
D. 42000 J ✓ Correct
Solution: Q = mcΔT = 2 × 4200 × 5 = 42000 J = 42 kJ.
Q11 — Specific Heat & Heat Capacity · easy · numerical
100 g of copper (c = 0.1 cal/g°C) is heated from 30°C to 80°C. The heat absorbed is:
A. 50 cal
B. 5000 cal
C. 500 cal ✓ Correct
D. 800 cal
Solution: Q = mcΔT = 100 × 0.1 × 50 = 500 cal.
Q12 — Specific Heat & Heat Capacity · easy · numerical
The heat capacity of 5 kg of a substance whose specific heat is 400 J/kg·K is:
A. 5000 J/K
B. 400 J/K
C. 80 J/K
D. 2000 J/K ✓ Correct
Solution: C = mc = 5 × 400 = 2000 J/K.
Q13 — Specific Heat & Heat Capacity · easy · numerical
A body of heat capacity 800 J/K absorbs 4000 J of heat. Its temperature rise is:
A. 8 K
B. 0.2 K
C. 5 K ✓ Correct
D. 2 K
Solution: ΔT = Q/C = 4000/800 = 5 K.
Q14 — Specific Heat & Heat Capacity · medium · numerical
The molar heat capacity of water (M = 18 g/mol, c = 1 cal/g°C) is:
A. 36 cal/mol°C
B. 1 cal/mol°C
C. 18 cal/mol°C ✓ Correct
D. 9 cal/mol°C
Solution: C_m = Mc = 18 × 1 = 18 cal/mol°C (≈ 75.4 J/mol·K).
Q15 — Specific Heat & Heat Capacity · medium · numerical
4200 J of work is completely converted into heat. The equivalent heat is (J = 4.2 J/cal):
A. 1000 cal ✓ Correct
B. 420 cal
C. 100 cal
D. 4200 cal
Solution: Q = W/J = 4200/4.2 = 1000 cal.
Q16 — Specific Heat & Heat Capacity · medium · numerical
Water falls from a height of 210 m. If all its energy becomes heat, the temperature rise is (g = 10 m/s², c = 4200 J/kg·K):
A. 0.05°C
B. 0.5°C ✓ Correct
C. 5°C
D. 2.1°C
Solution: ΔT = gh/c = (10 × 210)/4200 = 0.5°C. Speed-trick: ΔT ≈ h/420.
Q17 — Specific Heat & Heat Capacity · medium · numerical
Equal heat is supplied to 1 kg of water (c = 4200 J/kg·K) and 1 kg of oil (c = 2100 J/kg·K). If the water warms by 2°C, the oil warms by:
A. 4°C ✓ Correct
B. 8°C
C. 2°C
D. 1°C
Solution: Same Q and m: ΔT ∝ 1/c. Oil's c is half, so its rise is double = 4°C.
Q18 — Specific Heat & Heat Capacity · medium · numerical
A 2 kW electric heater warms 10 kg of water (c = 4200 J/kg·K) for 210 s. The temperature rise (no losses) is:
A. 21°C
B. 10°C ✓ Correct
C. 5°C
D. 42°C
Solution: Q = Pt = 2000 × 210 = 420000 J; ΔT = Q/mc = 420000/(10 × 4200) = 10°C.
Q19 — Specific Heat & Heat Capacity · medium · numerical
A 100 g copper ball (c = 0.1 cal/g°C) cools from 90°C to 40°C. The heat it releases is:
A. 900 cal
B. 500 cal ✓ Correct
C. 50 cal
D. 400 cal
Solution: Q = mcΔT = 100 × 0.1 × (90 − 40) = 500 cal released.
Q20 — Specific Heat & Heat Capacity · medium · numerical
A bullet moving at 200 m/s is stopped and 50% of its kinetic energy becomes heat within it (c = 500 J/kg·K). Its temperature rise is:
A. 40°C
B. 80°C
C. 20°C ✓ Correct
D. 10°C
Solution: Heat per kg = 0.5 × v²/2 = 0.5 × 20000 = 10000 J/kg; ΔT = 10000/500 = 20°C (mass cancels).