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Illustrations

Example-1

A plastic bucket contains 5 L of water at 20°C. One litre of hot water at 80°C is poured into it. Find the final temperature of water, given that the density of water is 1 kg L–1. Neglect the absorption of heat by the bucket and loss to the surrounding.

Solution

Let the final temperature be ‘t’ and the specific heat of water be ‘C’.

 

Mass of 5 L of water = 5 kg

 

Mass of 1 L of water = 1 kg.

 

Use Q = m × C  × ∆T

 

Heat lost by hot water, Q1 = 1 × C × (80 − t)

 

Heat gained by cold water,

 

Q2 = 5 × C  (t − 20).

 

By the principle of calorimetry, Q1 = Q2. Therefore,

 

C × (80 − t) = 5C(t − 20)

 

80 − t = 5t − 100

 

180 = 6t or t = 30°C.
 

 

Example-2

A hot iron ball of mass 0.2 kg is dropped into 2 kg of water at 20°C. The final temperature of water becomes 25°C. Calculate the initial temperature of iron.

 

Ciron = 480 J kg-1 K–1, Cwater = 4200 Jkg–1K–1.

Solution

Let the initial temperature of iron be ‘t’.

 

Use Q = m × C × ∆T

 

Heat lost by iron, Q1 = 0.2 × 480  (t − 25)

 

Heat gained by water,

 

Q2 = 2 × 4200 × (25 – 20)

 

By the principle of calorimetry, Q1 = Q2. Therefore,

 

0.2 × 480  (t − 25) = 2 × 4200 × 5

 

t − 25 = Description: 71931.png

 

t = (25 + 437.5) °C = 462.5°C
 

 

Example-3

A 1.5 kW water heater is used to heat 25 kg of water from 20°C to 40°C. Find the time taken assuming there are no losses. Cwater = 4200 J kg–1K–1.

Solution

Let the time taken be ‘t’ second.

 

Heat given by heater, Q1 = 1.5 ×103 × t Joules (1 Joule = 1 watt × 1 s)

 

Heat taken by water, Q2 = m × C × ∆T
= 25 × 4200 × (40 – 20) J
= 25 × 4200 × 20 = 21 × 105 J
Using Q1 = Q2, we have
1.5 × 103  t = 21 × 105
∴ t = 1400 s = 23 min 20 s.

 

 

Example-4

Earth receives solar energy at a rate of 1000 W m−2. A solar water heater of area 2 m2 absorbs 50% of incident energy. Find the time required to raise the temperature of 50 L of water by 50°C.

(Cw = 4180 J kg1 K1)

Solution

For 2 m2, rate = 2000 W

 

Total energy absorbed in a time t second is

 

Q1 = 2000  Description: 71945.png  t = (1000 × t) J

 

Heat required to raise the temperature of 50 L of water by 50°C is

 

Q2 = m Cw ∆T = 50 × 4180 × 50

 

= 1.045 × 107 J

 

We have, Q1 = Q2

 

∴ 1000t = 1.045 × 107

 

or t = Description: 71953.png = 10,450 s

 

or t = 2 h 54 min 10 s.
 


Example-5

A heat engine is cooled by 10 L of water. The engine has an efficiency of 50% and delivers 5 kW of mechanical power. Find the rise in temperature of water in 1 min (Cw = 4180 J kg1 K1).

Solution

Description: 71966.png

 

Description: 71980.png

 

Also Description: 71989.png

 

Description: 74895.png

 

∴ Q1 = 10 kW and Q2 = 5 kW.

 

This heat is rejected to water every second. Therefore, heat rejected by the engine into water in 1 min is Q = Q2  60 s = 300 kJ.

 

Let ∆T be the rise in temperature. Then,

Q = m  Cw  ∆T

Description: 71998.png

 

(The mass of 10 L of water is 10 kg.)
 





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