Grade 10 - Chap: 10 Heat Capacity & Modes of Transfer Solved Numericals Questions

Chapter 10: Heat Capacity and Modes of Transfer
Numerical Response Questions

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Q1. Convert the specific heat of water 4180 J/kg·K into units of J/g·°C.

Solution:

- Given:
  Specific heat = 4180 J/kg·K 
  We need to convert this into J/g·°C. 
- Concept: 
  1 kg = 1000 g and temperature difference in K is same as in °C (i.e., 1 K = 1°C).
- Conversion:
  4180 J/kg·K = 4180 J / 1000 g·°C = 4.18 J/g·°C

Answer: 4.18 J/g·°C

Q2. Calculate the amount of heat required to raise the temperature of 25 kg of water by 50°C.

Solution:

- Given: 
  Mass (m) = 25 kg 
  Temperature change (ΔT) = 50°C 
  Specific heat (c) = 4200 J/kg·K
- Formula: 
  Q = mcΔT
- Calculation:
  Q = 25 × 4200 × 50 = 5,250,000 J
- Convert to MJ:
  Q = 5.25 MJ

Answer: 5.25 MJ

Q3(a). A 0.5 kg block of unknown metal is heated from 30°C to 80°C using 19 kJ of heat. Calculate its specific heat capacity.

Solution:

- Given:
  Mass = 0.5 kg 
  Heat = 19 kJ = 19000 J  
  ΔT = 80°C - 30°C = 50°C
- Formula: 
  c = Q / (m × ΔT)
- Calculation:
  c = 19000 / (0.5 × 50) = 19000 / 25 = 760 J/kg·K

Answer: 760 J/kg·K

Q3(b). If the same metal block is further heated from 80°C to 125°C, how much more heat energy is required?

Solution:

- Use value of c from part a: 760 J/kg·K 
- ΔT = 125°C - 80°C = 45°C 
- Mass = 0.5 kg
- Formula:
  Q = mcΔT = 0.5 × 760 × 45 = 17100 J = 17.1 kJ

Answer: 17.1 kJ

Q4. Calculate the temperature change of 5 litres of water that absorbs 8.4 MJ of heat energy.

Solution:

- Volume = 5 L = 5 kg (since 1 L = 1 kg for water) 
- Q = 8.4 MJ = 8400000 J 
- c = 4200 J/kg·K
- Formula:
  ΔT = Q / (mc)
- Calculation:
  ΔT = 8400000 / (5 × 4200) = 8400000 / 21000 = 400°C

Answer: 400°C

Q5. Find the final temperature of mixture when 100 g water at 80°C is mixed with 200 g at 20°C.

Solution:

- Let final temp be Tf 
- No heat loss: heat lost by hot water = heat gained by cold water
- Fromula:
  m1cΔT1 = m2cΔT2
- Given:
  m1 = 100 g, T1 = 80°C 
  m2 = 200 g, T2 = 20°C 
  Specific heat cancels out (same for both).
Equation:
  100(80 - Tf) = 200(Tf - 20)
Expanding:
  8000 - 100Tf = 200Tf - 4000
Solving:
  12000 = 300Tf → Tf = 40°C

Answer: 40°C

Q6. A 200 g solid is heated to 110°C and placed in a calorimeter with 200 g water at 20°C and 60 g calorimeter. Final temperature is 35°C. Find the specific heat of the solid.

Solution:

- Given:
  Solid: m = 200 g, T_initial = 110°C 
  Water: m = 200 g, c = 4.2 J/g°C, T_initial = 20°C 
  Calorimeter: m = 60 g, c = 0.9 J/g°C 
  Final temperature = 35°C
Let c(s) = specific heat of solid
- Heat lost by solid:
  Q(s) = m × c(s) × ΔT = 200 × c(s) × (110 - 35) = 200 × c(s) × 75
- Heat gained by water:
  Q(w) = 200 × 4.2 × (35 - 20) = 12600 J
- Heat gained by calorimeter:
  Q(c) = 60 × 0.9 × (35 - 20) = 810 J
Total heat gained:
  Q(total) = 12600 + 810 = 13410 J
Equation:
  200 × c(s) × 75 = 13410 → c(s) = 13410 / 15000 = 0.894 J/g°C

Answer: 0.89 J/g°C or 890 J/kg·K

Q7. A steel rod of mass 3 kg cools from 450°C to 50°C. Find the heat lost. Specific heat = 460 J/kg·K

Solution:

- Given:
  Mass = 3 kg 
  ΔT = 450°C - 50°C = 400°C 
  c = 460 J/kg·K
- Formula:
  Q = mcΔT = 3 × 460 × 400 = 552000 J = 552 kJ

Answer: 552 kJ