HC Verma Solutions: Chapter 25 - Calorimetry | Physics Class 11 - NEET PDF Download

 Page 1


25.1
CHAPTER – 25
CALORIMETRY
1. Mass of aluminium = 0.5kg, Mass of water = 0.2 kg
Mass of Iron = 0.2 kg Temp. of aluminium and water = 20°C = 297°k
Sp heat o f Iron = 100°C = 373°k. Sp heat of aluminium = 910J/kg-k
Sp heat of Iron = 470J/kg-k Sp heat of water = 4200J/kg-k
Heat again = 0.5 × 910(T – 293) + 0.2 × 4200 × (343 –T)
= (T – 292) (0.5 × 910 + 0.2 × 4200) Heat lost = 0.2 × 470 × (373 – T)
? Heat gain   = Heat lost
? (T – 292) (0.5 × 910 + 0.2 × 4200) = 0.2 × 470 × (373 – T)
? (T – 293) (455 + 8400)  = 49(373 – T)
? (T – 293) ?
?
?
?
?
?
94
1295
= (373 – T)
?  (T – 293) × 14 = 373 – T 
? T = 
15
4475
= 298 k 
? T = 298 – 273 = 25°C. The final temp = 25°C.
2. mass of Iron = 100g water Eq of caloriemeter = 10g
mass of water = 240g Let the Temp. of surface = 0 ?C
S
iron
= 470J/kg°C Total heat gained = Total heat lost.
So, 
1000
100
× 470 ×( ? – 60) = 
1000
250
× 4200 × (60 – 20)
? 47 ? – 47 × 60 = 25 × 42 × 40
? ? = 4200 + 
47
2820
= 
47
44820
= 953.61°C ?
3. The temp. of A = 12°C The temp. of B = 19°C
The temp. of C = 28°C The temp of ? A + B = 16°
The temp. of ? B + C = 23°
In accordance with the principle of caloriemetry when A & B are mixed
M
CA
(16 – 12) = M
CB
(19 – 16) ? CA4 = CB3  ? CA = 
4
3
CB …(1)
And when B & C are mixed
M
CB
(23 – 19)= M
CC
(28 – 23) ? 4CB = 5CC ? CC = 
5
4
CB …(2)
When A & c are mixed, if T is the common temperature of mixture
M
CA
(T – 12) = M
CC
(28 – T) 
? ?
?
?
?
?
?
4
3
CB(T – 12) = ?
?
?
?
?
?
5
4
CB(28 – T) 
? 15T – 180 = 448 – 16T 
? T = 
31
628
= 20.258°C = 20.3°C 
? ? ? ? ??