What is the mass of a material, whose specific heat capacity is 400 J/(kg °C) for a rise in temperature from 15 °C to 25 °C, when heat received is 20 kJ?
- (a)0·1 kg
- (b)1 kg
- (c)10 kg
- (d)5 kg
Correct — D, 5 kg. Heat absorbed follows Q = m·c·ΔT, so the mass is m = Q / (c·ΔT). Here Q = 20 kJ = 20 000 J, the specific heat capacity c = 400 J/(kg·°C) and the temperature rise ΔT = 25 °C − 15 °C = 10 °C. Substituting, m = 20 000 / (400 × 10) = 20 000 / 4 000 = 5 kg.
- (a)0·1 kg — This comes from a stray division (for example dividing 400 by the heat, or forgetting to convert kilojoules to joules); the correct m = Q/(cΔT) = 20 000/4 000 = 5 kg.
- (b)1 kg — Using the final temperature (25 °C) as ΔT and not converting 20 kJ to joules gives roughly 1 kg; ΔT must be the rise 25 − 15 = 10 °C and Q = 20 000 J, giving 5 kg.
- (c)10 kg — This mis-multiplies the denominator (for example using c·ΔT = 2 000 instead of 4 000); the arithmetic 20 000/(400 × 10) is exactly 5 kg, not 10 kg.
Specific heat capacity (c) is the amount of heat needed to raise the temperature of 1 kg of a substance by 1 °C. The heat exchanged when a body of mass m warms or cools by ΔT is given by Q = m·c·ΔT, which can be rearranged to find any one of the four quantities.
This is a plug-in use of Q = mcΔT. The only traps are unit conversion (20 kJ = 20 000 J) and using the temperature rise (10 °C) rather than the final temperature (25 °C). Rearrange to m = Q/(cΔT) and the answer falls out.
- Specific heat capacity here is c = 400 J/(kg·°C) — the heat to warm 1 kg by 1 °C.
- The governing relation is Q = m·c·ΔT, so m = Q/(c·ΔT).
- ΔT is the temperature change (25 − 15 = 10 °C), not the final temperature.
- Units must match: 20 kJ = 20 000 J before dividing.
- Forgetting to convert kilojoules to joules before dividing.
- Using the final temperature (25 °C) instead of the temperature rise (10 °C) for ΔT.
A direct numerical use of Q = mcΔT — the only twists are converting kJ to J and using the temperature rise, not the final temperature.
UPSC_1999_GS1_Q1281999Assertion (A): To dilute sulphuric acid, acid is added to water and not water to acid. Reason (R): Specific heat of water is quite large.
- (a) Both A and R are true, and R is the correct explanation of A
- (b) Both A and R are true, but R is not a correct explanation of A
- (c) A is true, but R is false
- (d) A is false, but R is true
Answer(a) Both A and R are true, and R is the correct explanation of A
Same concept — the specific heat capacity of water. That UPSC item leans on water's large specific heat (it absorbs a lot of heat per degree); this one uses the specific-heat relation Q = mcΔT to find a mass.
10 g of ice at -10°C is mixed with 10 g of water at 0°C. The amount of heat required to raise the temperature of mixture to 10°C is
- (a) 400 cal
- (b) 550 cal
- (c) 1050 cal
- (d) 1200 cal
Answer(c) 1050 cal
Same area — calorimetry using Q = mcΔT (with latent heat added). That NDA item sums the heat to warm and melt a mixture; this one applies the same Q = mcΔT relation to find a mass.
- practice — not a real PYQ
A body of mass 2 kg is heated from 20 °C to 45 °C, absorbing 25 000 J of heat. Its specific heat capacity is
- (a)250 J/(kg·°C)
- (b)400 J/(kg·°C)
- (c)500 J/(kg·°C)
- (d)1000 J/(kg·°C)
Answer(c) 500 J/(kg·°C) — c = Q/(mΔT) = 25 000/(2 × 25).
- practice — not a real PYQ
How much heat is needed to raise the temperature of 3 kg of a substance of specific heat 200 J/(kg·°C) by 20 °C?
- (a)1 200 J
- (b)6 000 J
- (c)12 000 J
- (d)60 000 J
Answer(c) 12 000 J — Q = mcΔT = 3 × 200 × 20.