Science > Physics > Thermometry > Numerical Problems on Principle of Heat Exchange
In the last article, we have studied out calorimetry. In this article, we shall study more numerical problems on the principle of heat exchange.
Law of Heat Exchange:
When a hot body is mixed or kept in contact with a cold body, the hot body loses heat and its temperature falls. On the other hand, the cold body absorbs the heat lost by the hot body and the temperature of the cold body increases. A state will reach when the whole mixture shows a common temperature. This final temperature of the mixture will lie between the original temperatures of the hot body and the cold body. If this system is completely isolated that is there is no flow of energy into the system or out of the system, then by the law of conservation of energy, the heat lost by one body is equal to the heat gained by the other body. This relation is known as the law of heat exchange.
Thus by the law of heat exchange, Heat lost = Heat gained.
Numerical Problems:
Example 01:
What will be the final temperature of the mixture when 10 g of ice at – 5 oC mixed with 60 g of water at 20 oC. Latent heat of fusion of ice = 80 cal/g and specific heat of ice is 0.5 cal/g/oC.
Given: Mass of ice mi = 10 g, Temperature of ice Ti = – 5 oC, Specific heat of ice = ci = 0.5 cal/g/oC, Temperature of water = Tw = 20 oC, Mass of water = mw = 60 g, Latent heat of fusion of ice = Li = 80 cal/g.
To Find: Final temperature of mixture = T =?
Solution:
Heat gained by ice = mi ci (0 – Ti) + miLi + micw(T – 0)
∴ Heat gained by ice = 10 x 0.5 x (0 – (-5)) + 10 x 80 + + 10 x 1 x(T – 0)
∴ Heat gained by ice = 10 x 0.5 x 5 + 800 + 10 T
∴ Heat gained by ice = (825 + 10 T) cal ………… (1)
Heat lost by water = mw cw (Tw – T)
∴ Heat lost by water = 60 x 1 x (20 – T)
∴ Heat lost by water = (1200 – 60T) cal ………… (2)
By the principle of heat exchange
Heat gained by ice = Heat lost by water
∴ 825 + 10 T = 1200 – 60T
∴ 70 T = 375
∴ T = 375/70 = 5.36 oC
Ans: Final temperature of mixture is 5.36 oC
Example 02:
A piece of brass is heated to a temperature of 120 and dropped into a cavity in a block of ice at 0 . If the sp.ht of brass is 0.1 cal/g/ and the mass of brass taken is 15 g, find the mass of ice melted. ( L of fusion of ice = 80 cal/g)
Given: Temperature of ice Ti = 0 oC, Temperature of brass piece = Tb = 120 oC, Mass of brass piece = mb = 15 g, Sp. Heat of brass = cb = 0.1 cal/g/ , Latent heat of fusion of ice = Li = 80 cal/g.
To Find: Mass of ice = mi =?
Solution:
As ice is still melting, temperature of mixture = T = 0 oC
Heat gained by ice = miLi = mi x 80
Heat gained by ice = 80 mi cal ………… (1)
Heat lost by brass piece = mb cb (Tb – T)
∴ Heat lost by brass piece = 15 x 0.1 x (120 – 0)
∴ Heat lost by brass piece = 180 cal ………… (2)
By the principle of heat exchange
Heat gained by ice = Heat lost by brass piece
∴ 80 mi = 180
∴ mi = 180/80 = 2.25 g
Ans: Thus mass of ice melted is 2.25 g
Example 03:
A piece of iron weighing 15 g is dropped into a cavity in a block of ice 0 oC, 2.5 g of ice are melted. If the temperature of iron was 113.6 oC, calculate the Sp.ht of iron. (L for the fusion of ice = 80 cal/g)
Given: Mass of ice melted = mi = 2.5 g, Temperature of ice Ti = 0 oC, Temperature of iron piece = TIron = 113.6 oC, Mass of iron piece = miron = 15 g, Latent heat of fusion of ice = Li = 80 cal/g.
To Find: Sp. Heat of iron = cIron =?
Solution:
As ice is still melting, temperature of mixture = T = 0 oC
Heat gained by ice = miLi = 2.5 x 80 = 200 cal ………… (1)
Heat lost by iron piece = mIron cIron (TIron – T)
∴ Heat lost by iron piece = 15 x cIron x (113.6 – 0)
∴ Heat lost by iron piece = 1704 cIron cal ………… (2)
By the principle of heat exchange
Heat gained by ice = Heat lost by iron piece
∴ 200 = 1704 cIron
∴ cIron = 200/1704 = 0.1174 = 0.12 cal/g/oC
Ans: Thus specific heat of iron is 0.12 cal/g/oC
Example 04:
A copper calorimeter of mass 50 g contains 500 g of water at 30 oC. A piece of ice of mass 10 g and at a temperature -5 oC is dropped into the calorimeter. Find the final temperature of the calorimeter and it contains when all the ice has melted. (Sp. Heat of copper = 0.09, sp.ht of ice = 0.5 and latent heat of ice = 80 cal/g)
Given: Mass of calorimeter = mC = 50g, Specific heat of copper = 0.09 cal/g/oC, Mass of water mW = 500 g, Initial temperature of water and the calorimeter = TW = 30oC, Mass of ice mi = 10 g, Temperature of ice Ti = – 5 oC, Specific heat of ice = ci = 0.5 cal/g/oC, Latent heat of fusion of ice = Li = 80 cal/g.
To Find: Final temperature of mixture = T =?
Solution:
Heat gained by ice = mi ci (0 – Ti) + miLi + micw(T – 0)
∴ Heat gained by ice = 10 x 0.5 x (0 – (-5)) + 10 x 80 + 10 x 1 x(T – 0)
∴ Heat gained by ice = 10 x 0.5 x 5 + 800 + 10T
∴ Heat gained by ice = (825 + 10 T) cal ………… (1)
Heat lost by water and calorimeter = (mccc + mw) cw (Tw – T)
∴ Heat lost by water and calorimeter = (50 x 0.09 + 500) x 1 x (30 – T)
∴ Heat lost by water and calorimeter = (4.5 + 500) x (30 – T)
∴ Heat lost by water and calorimeter = 504.5 x (30 – T) cal ………… (2)
By the principle of heat exchange
Heat gained by ice = Heat lost by water and calorimeter
∴ 825 + 10T = 504.5 x (30 – T)
∴ 825 + 10T = 15140 – 504.5T
∴ 10T + 504.5T = 15140 – 825
∴ 514.5 T = 14310
∴ T = 14310/514.5 = 27.81 oC
Ans: The final temperature of the calorimeter and it contains when all the ice has melted is 27.81 oC
Example 05:
When a piece of metal of mass 48.3 g at 10.7 oC was immersed in a current of steam at 100 oC, 0.762 g of steam was found to condense. Calculate the specific heat of the metal.
Given: Mass of steam = ms = 0.762 g, Temperature of steam Ts = 100 oC, Temperature of metal piece = Tm = 10.7 oC, Mass of metal piece = mm = 48.3 g.
To Find: Sp. Heat of metal = cm =?
Solution:
Heat lost by steam = miLi = 0.762 x 540 cal ………… (1)
Heat gained by metal piece = mm cm (T – Tm)
∴ Heat gained by metal piece = 48.3 x cm x (100 – 10.7)
∴ Heat gained by metal piece = 48.3 x 89.3 x cm cal ………… (2)
By the principle of heat exchange
Heat lost by steam = Heat gained by metal piece
∴ 0.762 x 540 = 48.3 x 89.3 x cm
∴ cm = (0.762 x 540)/( 48.3 x 89.3) = 0.0954 = 0.095 cal/g/oC
Ans: Thus specific heat of metal is 0.095 cal/g/oC