Specific heat and calorimetry: Class 11 practice
Heat needed Q = mcΔT. Read the method, follow the three worked examples, then try the six questions.
What you should know first
- Linear equations
The idea behind specific heat and calorimetry
Heat needed Q = mcΔT. When hot and cold bodies mix with no losses, heat lost = heat gained, which gives the final temperature.
Step-by-step method
- Write m, c and the temperature change for each body.
- Heat lost by hot = heat gained by cold.
- Solve for the unknown.
Worked example 1 (Basic)
How much heat raises 5 kg of water by 11 °C? (c = 4200 J/kg K)
- Q = mcΔT = 5 × 4200 × 11
- = 231000 J
- Answer: 231000 J
Worked example 2 (Standard)
4 kg of water at 67 °C is mixed with 2 kg at 23 °C. Find the final temperature (no heat loss).
- 4(67 − T) = 2(T − 23)
- T = 52.33 °C
- Answer: 52.33 °C
Worked example 3 (Challenge)
3 kg of water at 67 °C is mixed with 3 kg at 5 °C. Find the final temperature (no heat loss).
- 3(67 − T) = 3(T − 5)
- T = 36 °C
- Answer: 36 °C
Common mistakes and misconceptions
- Mixing grams and kg with c in J/kg K
- Wrong sign for temperature change
Practice questions
- (Basic) How much heat raises 5 kg of water by 52 °C? (c = 4200 J/kg K)
- (Basic) How much heat raises 4 kg of water by 23 °C? (c = 4200 J/kg K)
- (Standard) 5 kg of water at 62 °C is mixed with 4 kg at 26 °C. Find the final temperature (no heat loss).
- (Standard) 1 kg of water at 60 °C is mixed with 2 kg at 27 °C. Find the final temperature (no heat loss).
- (Challenge) 3 kg of water at 69 °C is mixed with 2 kg at 24 °C. Find the final temperature (no heat loss).
- (Challenge) 3 kg of water at 83 °C is mixed with 2 kg at 27 °C. Find the final temperature (no heat loss).
Answer key with full working
- Q1: Q = mcΔT = 5 × 4200 × 52 → = 1092000 J → Answer: 1092000 J
- Q2: Q = mcΔT = 4 × 4200 × 23 → = 386400 J → Answer: 386400 J
- Q3: 5(62 − T) = 4(T − 26) → T = 46 °C → Answer: 46 °C
- Q4: 1(60 − T) = 2(T − 27) → T = 38 °C → Answer: 38 °C
- Q5: 3(69 − T) = 2(T − 24) → T = 51 °C → Answer: 51 °C
- Q6: 3(83 − T) = 2(T − 27) → T = 60.6 °C → Answer: 60.6 °C
Syllabus note
This skill is listed under "Thermal Properties of Matter" for Class 11 in the NCERT/CBSE books. ICSE, IGCSE, IB and state boards may teach it in a different chapter or year.
Answers on this page are generated and automatically re-checked by a second, independent calculation. They have not been reviewed by a teacher — if you spot a problem, tell us on WhatsApp.
Related practice and notes
Frequently asked questions
Which chapter is specific heat and calorimetry in?
"Thermal Properties of Matter", Class 11 Physics (NCERT/CBSE). Other boards may place it differently.
How were the answers checked?
Answers on this page are generated and automatically re-checked by a second, independent calculation. They have not been reviewed by a teacher — if you spot a problem, tell us on WhatsApp.