Rate of heat conduction: Class 11 practice
Heat flow through a slab: H = kA(T₁ − T₂)/L, where k is thermal conductivity, A the area and L the thickness. Read the method, follow the three worked examples, then try the six questions.
What you should know first
- Units
The idea behind rate of heat conduction
Heat flow through a slab: H = kA(T₁ − T₂)/L, where k is thermal conductivity, A the area and L the thickness. Thicker slabs or smaller areas reduce heat flow.
Step-by-step method
- Convert area to m² and thickness to m.
- Temperature difference in K or °C (same).
- H = kAΔT/L, in watts.
Worked example 1 (Basic)
A slab (k = 50 W/m K) of area 0.08 m² and thickness 6 cm has faces at 89 °C and 22 °C. Find the heat flow rate.
- H = 50 × 0.08 × 67/0.06
- Answer: 4466.67 W
Worked example 2 (Standard)
A slab (k = 200 W/m K) of area 0.2 m² and thickness 7.000000000000001 cm has faces at 74 °C and 1 °C. Find the heat flow rate.
- H = 200 × 0.2 × 73/0.07
- Answer: 41714.29 W
Worked example 3 (Challenge)
How much heat crosses that slab (H = 2280 W) in 9 minutes?
- Q = H × t = 2280 × 540
- Answer: 1231200 J
Common mistakes and misconceptions
- Leaving thickness in cm
- Using one temperature instead of the difference
Practice questions
- (Basic) A slab (k = 50 W/m K) of area 0.18 m² and thickness 9 cm has faces at 76 °C and 30 °C. Find the heat flow rate.
- (Basic) A slab (k = 0.8 W/m K) of area 0.08 m² and thickness 6 cm has faces at 83 °C and 18 °C. Find the heat flow rate.
- (Standard) A slab (k = 50 W/m K) of area 0.08 m² and thickness 7.000000000000001 cm has faces at 87 °C and 12 °C. Find the heat flow rate.
- (Standard) A slab (k = 1.2 W/m K) of area 0.08 m² and thickness 1 cm has faces at 84 °C and 17 °C. Find the heat flow rate.
- (Challenge) How much heat crosses that slab (H = 23600 W) in 7 minutes?
- (Challenge) How much heat crosses that slab (H = 9900 W) in 7 minutes?
Answer key with full working
- Q1: H = 50 × 0.18 × 46/0.09 → Answer: 4600 W
- Q2: H = 0.8 × 0.08 × 65/0.06 → Answer: 69.33 W
- Q3: H = 50 × 0.08 × 75/0.07 → Answer: 4285.71 W
- Q4: H = 1.2 × 0.08 × 67/0.01 → Answer: 643.2 W
- Q5: Q = H × t = 23600 × 420 → Answer: 9912000 J
- Q6: Q = H × t = 9900 × 420 → Answer: 4158000 J
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 rate of heat conduction 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.