# Illustrations

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

Let the final temperature be â€˜tâ€™ and the specific heat of water be â€˜Câ€™.

Mass of 5 L of water = 5 kg

Mass of 1 L of water = 1 kg.

Use Q = m Ã— C Ã— âˆ†T

Heat lost by hot water, Q_{1} = 1 Ã— C Ã— (80 âˆ’ t)

Heat gained by cold water,

Q_{2} = 5 Ã— C (t âˆ’ 20).

By the principle of calorimetry, Q_{1} = Q_{2}. Therefore,

C Ã— (80 âˆ’ t) = 5C(t âˆ’ 20)

80 âˆ’ t = 5t âˆ’ 100

180 = 6t or t = 30Â°C.

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

C_{iron} = 480 J kg^{-1} K^{â€“1}, C_{water }= 4200 Jkg^{â€“1}K^{â€“1}.

Let the initial temperature of iron be â€˜tâ€™.

Use Q = m Ã— C Ã— âˆ†T

Heat lost by iron, Q_{1} = 0.2 Ã— 480 (t âˆ’ 25)

Heat gained by water,

Q_{2} = 2 Ã— 4200 Ã— (25 â€“ 20)

By the principle of calorimetry, Q_{1} = Q_{2}. Therefore,

0.2 Ã— 480 (t âˆ’ 25) = 2 Ã— 4200 Ã— 5

t âˆ’ 25 =

t = (25 + 437.5) Â°C = 462.5Â°C

A 1.5 kW water heater is used to heat 25 kg of water from 20Â°C to 40Â°C. Find the time taken assuming there are no losses. C_{water} = 4200 J kg^{â€“1}K^{â€“1}.

Let the time taken be â€˜tâ€™ second.

Heat given by heater, Q_{1} = 1.5 Ã—10^{3} Ã— *t* Joules (1 Joule = 1 watt Ã— 1 s)

Heat taken by water, Q_{2} = m Ã— C Ã— âˆ†T

= 25 Ã— 4200 Ã— (40 â€“ 20) J

= 25 Ã— 4200 Ã— 20 = 21 Ã— 10^{5} J

Using Q_{1} = Q_{2}, we have

1.5 Ã— 10^{3} t = 21 Ã— 10^{5}

âˆ´ t = 1400 s = 23 min 20 s.

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

(C_{w} = 4180 J kg^{âˆ’}^{1} K^{âˆ’}^{1})

For 2 m^{2}, rate = 2000 W

Total energy absorbed in a time t second is

Q_{1} = 2000 t = (1000 Ã— t) J

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

Q_{2} = m C_{w} âˆ†T = 50 Ã— 4180 Ã— 50

= 1.045 Ã— 10^{7} J

We have, Q_{1} = Q_{2}

âˆ´ 1000t = 1.045 Ã— 10^{7}

or t = = 10,450 s

or t = 2 h 54 min 10 s.

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 (C_{w} = 4180 J kg^{âˆ’}^{1} K^{âˆ’}^{1}).

Also

âˆ´ Q_{1} = 10 kW and Q_{2} = 5 kW.

This heat is rejected to water every second. Therefore, heat rejected by the engine into water in 1 min is Q = Q_{2} 60 s = 300 kJ.

Let âˆ†T be the rise in temperature. Then,

Q = m C_{w} âˆ†T

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