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Solved Examples on Heat Phenomena:- Question 1:- (a) Calculate the rate of heat loss through a glass window of area 1.4 m^{2} and thickness 3.0 mm if the outside temperature is -20º F and the inside temperature is +72º F. (b) A storm window is installed having the same thickness of glass but with an air gap of 7.5 cm between the two windows. What will be the corresponding rate of heat loss presuming that conduction is the only important heat-loss mechanism? Concept:- The rate H at which heat is transferred through the rod is, (a) directly proportional to the cross-sectional area (A) available. (b) inversely proportional to the length of the rod Δx. (c) directly proportional to the temperature difference ΔT. So, H = kA ΔT/ Δx Here k is the proportionality constant and is called thermal conductivity of the rod. Again the rate of heat transfer H is defined as, H = A(T_{2}-T_{1})/∑R_{n} = A ΔT/ (R_{g}+ R_{a}) Here A is the area, ΔT is the temperature difference, R_{g} is the thermal resistance of glass at the thickness 3.0 mm and R_{a} is the thermal resistance of air at the thickness 7.5 cm. The thermal resistance or R-value, defined by, R = L/k Here L is the thickness of the material through which the heat is transferred and k is the thermal conductivity of the material. Solution:- (a) First we have to find the temperature difference between inside and outside of the glass window. If T_{2} is the inside temperature and T_{1} is the outside temperature, then the temperature difference ΔT in ^{°}C will be, ΔT = T_{2}-T_{1} = 5^{°}C /9 ^{°}F (72^{°}F –(-20 ^{° }F)) = 51.1 ^{°}C = (51.1+273) K = 324.1 K To obtain the rate of heat loss through a glass window, substitute 1.0 W/m.K (thermal conductivity of window glass) for k, 1.4 m^{2} for A, 324.1 K for ΔT and 3.0 mm for Δx in the equation H = kA ΔT/ Δx, we get, H = kA ΔT/ Δx = (1.0 W/m.K) (1.4 m^{2}) (324.1 K)/ (3.0 mm) = (1.0 W/m.K) (1.4 m^{2}) (324.1 K)/ (3.0 mm×10^{-3} m/1 mm) = 2.4×10^{4} W From the above observation we conclude that, the rate of heat loss through a glass window would be 2.4×10^{4} W. (b) First we have to find out the thermal resistance R_{g} of galss at the thickness 3.0 mm and R_{a} of air at the thickness 7.5 cm. To obtain the thermal resistance R_{g} of galss at the thickness 3.0 mm, substitute 3.0 mm for L and 1.0 W/m.K (thermal conductivity of window glass) for k in the equation R = L/k, R_{g} = L/k = 3.0 mm/1.0 W/m.K = (3.0 mm×10^{-3} m/1 mm) /(1.0 W/m.K) = 3.0×10^{-3} m^{2}.K/W To obtain the thermal resistance R_{a} of air at the thickness 7.5 cm, substitute 7.5 cm for L and 0.026 W/m.K (thermal conductivity of dry air) for k in the equation R = L/k, R_{a} = L/k = 7.5 cm /0.026 W/m.K = (7.5 cm ×10^{-2} m/1 mm) /(0.026 W/m.K) = 2.88 m^{2}.K/W To find out the corresponding rate of heat loss H, substitute 1.4 m^{2} for A,3.0×10^{-3} m^{2}.K/W for R_{g} and 2.88 m^{2}.K/W for R_{a} in the equation H = A ΔT/ (R_{g}+ R_{a}), we get, H = A ΔT/ (R_{g}+ R_{a}) = (1.4 m^{2}) (324.1 K) /(3.0×10^{-3} m^{2}.K/W) (2.88 m^{2}.K/W) = 25 W From the above observation we conclude that, the corresponding rate of heat loss H would be 25 W. ______________________________________________________________________________________________ Question 2:- What mass of steam at 100ºC must be mixed with 150 g of ice at 0ºC, in a thermally insulated container, to produce liquid water at 50ºC? Concept:- The heat capacity per unit mass of a body, called specific heat capacity or usually just specific heat, is characteristic of the material of which the body is composed. c = C/m = Q/mΔT So, Q = c mΔT Here, the heat transferred is Q, specific heat capacity is c, mass is m and the temperature difference is ΔT. The amount of heat per unit mass that must be transferred to produce a phase change is called the heat of transformation or latent heat L for the process. The total heat Q transferred in a phase change is then, Q = Lm Here m is the mass of the sample that changes phase. Solution:- The heat given off the steam Q_{s} will be equal to, Q_{s} = m_{s}L_{v}+ m_{s}c_{w}ΔT Here, mass of steam is m_{s}, latent heat vaporization is L_{v}, specific heat capacity of water is c_{w} and the temperature difference is ΔT. The heat taken in by the ice Q_{i} will be equal to, Q_{i} = m_{i}L_{f}+ m_{i}c_{w}ΔT Here, mass of ice is m_{i}, latent heat fusion is L_{f}, specific heat capacity of water is c_{w} and the temperature difference is ΔT. Heat given off the steam Q_{s} is equal to the heat taken in by the ice Q_{i}. So, Q_{s} = Q_{i} m_{s}L_{v}+ m_{s}c_{w}ΔT = m_{i}L_{f}+ m_{i}c_{w}ΔT m_{s}(L_{v}+ c_{w}ΔT) = m_{i}(L_{f}+ c_{w}ΔT) m_{s} = m_{i}(L_{f}+ c_{w}ΔT)/ (L_{v}+ c_{w}ΔT) To obtain the mass of the steam at 100^{ °}C must be mixed with 150 g of ice at 0 ^{°}C, substitute 150 g for mass of ice m_{i}, 333×10^{3} J/kg for L_{f}, 4190 J/kg.K for c_{w}, 50^{°} C for ΔT, 2256×10^{3} J/kg for L_{v} in the equation m_{s} = m_{i}(L_{f}+ c_{w}ΔT)/ (L_{v}+ c_{w}ΔT), we get, m_{s} = m_{i}(L_{f}+ c_{w}ΔT)/ (L_{v}+ c_{w}ΔT) =(150 g)[(333×10^{3} J/kg) +(4190 J/kg.K) (50^{°} C)]/ [(2256×10^{3} J/kg)+ (4190 J/kg.K) (50^{°} C)] =(150 g×(10^{-3} kg/1g))[(333×10^{3} J/kg) +(4190 J/kg.K) (50+273)K]/ [(2256×10^{3} J/kg)+ (4190 J/kg.K) (50+273)K] = 0.033 kg From the above observation we conclude that, the mass of steam at 100^{ °}C must be mixed with 150 g of ice at 0 ^{°}C would be 0.033 kg. _____________________________________________________________________________________________________ Question 3:- A small electric immersion heater is used to boil 136 g of water for a cup of instant coffee. The heater is labeled 220 watts. Calculate the time required to bring this water from 23.5ºC to the boiling point, ignoring nay heat losses. Concept:- Heat Q that must be given to a body of mass m, whose material has a specific heat c, to increase its temperature from initial temperature T_{i} to final temperature T_{f} is, Q = mc (T_{f} - T_{i}) But time (t) is equal to the heat energy (Q) divided by power (P). t = Q/ P = mc (T_{f} - T_{i})/P Solution:- To obtain the time required to bring this water from 23.5^{°} C to the boiling point, substitute 136 g for mass of water m, 4190 J/kg. K for specific heat capacity of water c, 100^{°} C for final temperature T_{f} (boiling point of water), 23.5° C for initial temperature T_{i} and 220 watts for power P in the equation t = mc (T_{f} - T_{i})/P, t = mc (T_{f} - T_{i})/P = (136 g) (4190 J/kg. K) (100^{°} C - 23.5° C) / 220 W = (136 g×10^{-3} kg/1 g) (4190 J/kg. K) (100^{°} C - 23.5° C) / 220 W = (0.136 kg) (4190 J/kg. K) ((100+273) K – (23.5 + 273)K) / 220 W = (0.136 kg) (4190 J/kg. K) (373 K – 296.5 K) / 220 W = (0.136 kg) (4190 J/kg. K) (76.5 K) / 220 W = 198.15 s Rounding off to three significant figures, the time required to bring this water from 23.5^{°} C to the boiling point would be 198 s.
(a) Calculate the rate of heat loss through a glass window of area 1.4 m^{2} and thickness 3.0 mm if the outside temperature is -20º F and the inside temperature is +72º F.
(b) A storm window is installed having the same thickness of glass but with an air gap of 7.5 cm between the two windows. What will be the corresponding rate of heat loss presuming that conduction is the only important heat-loss mechanism?
The rate H at which heat is transferred through the rod is,
(a) directly proportional to the cross-sectional area (A) available.
(b) inversely proportional to the length of the rod Δx.
(c) directly proportional to the temperature difference ΔT.
So, H = kA ΔT/ Δx
Here k is the proportionality constant and is called thermal conductivity of the rod.
Again the rate of heat transfer H is defined as,
H = A(T_{2}-T_{1})/∑R_{n}
= A ΔT/ (R_{g}+ R_{a})
Here A is the area, ΔT is the temperature difference, R_{g} is the thermal resistance of glass at the thickness 3.0 mm and R_{a} is the thermal resistance of air at the thickness 7.5 cm.
The thermal resistance or R-value, defined by,
R = L/k
Here L is the thickness of the material through which the heat is transferred and k is the thermal conductivity of the material.
(a) First we have to find the temperature difference between inside and outside of the glass window.
If T_{2} is the inside temperature and T_{1} is the outside temperature, then the temperature difference ΔT in ^{°}C will be,
ΔT = T_{2}-T_{1}
= 5^{°}C /9 ^{°}F (72^{°}F –(-20 ^{° }F))
= 51.1 ^{°}C
= (51.1+273) K
= 324.1 K
To obtain the rate of heat loss through a glass window, substitute 1.0 W/m.K (thermal conductivity of window glass) for k, 1.4 m^{2} for A, 324.1 K for ΔT and 3.0 mm for Δx in the equation H = kA ΔT/ Δx, we get,
H = kA ΔT/ Δx
= (1.0 W/m.K) (1.4 m^{2}) (324.1 K)/ (3.0 mm)
= (1.0 W/m.K) (1.4 m^{2}) (324.1 K)/ (3.0 mm×10^{-3} m/1 mm)
= 2.4×10^{4} W
From the above observation we conclude that, the rate of heat loss through a glass window would be 2.4×10^{4} W.
(b) First we have to find out the thermal resistance R_{g} of galss at the thickness 3.0 mm and R_{a} of air at the thickness 7.5 cm.
To obtain the thermal resistance R_{g} of galss at the thickness 3.0 mm, substitute 3.0 mm for L and 1.0 W/m.K (thermal conductivity of window glass) for k in the equation R = L/k,
R_{g} = L/k
= 3.0 mm/1.0 W/m.K
= (3.0 mm×10^{-3} m/1 mm) /(1.0 W/m.K)
= 3.0×10^{-3} m^{2}.K/W
To obtain the thermal resistance R_{a} of air at the thickness 7.5 cm, substitute 7.5 cm for L and 0.026 W/m.K (thermal conductivity of dry air) for k in the equation R = L/k,
R_{a} = L/k
= 7.5 cm /0.026 W/m.K
= (7.5 cm ×10^{-2} m/1 mm) /(0.026 W/m.K)
= 2.88 m^{2}.K/W
To find out the corresponding rate of heat loss H, substitute 1.4 m^{2} for A,3.0×10^{-3} m^{2}.K/W for R_{g} and 2.88 m^{2}.K/W for R_{a} in the equation H = A ΔT/ (R_{g}+ R_{a}), we get,
H = A ΔT/ (R_{g}+ R_{a})
= (1.4 m^{2}) (324.1 K) /(3.0×10^{-3} m^{2}.K/W) (2.88 m^{2}.K/W)
= 25 W
From the above observation we conclude that, the corresponding rate of heat loss H would be 25 W.
______________________________________________________________________________________________
What mass of steam at 100ºC must be mixed with 150 g of ice at 0ºC, in a thermally insulated container, to produce liquid water at 50ºC?
The heat capacity per unit mass of a body, called specific heat capacity or usually just specific heat, is characteristic of the material of which the body is composed.
c = C/m
= Q/mΔT
So, Q = c mΔT
Here, the heat transferred is Q, specific heat capacity is c, mass is m and the temperature difference is ΔT.
The amount of heat per unit mass that must be transferred to produce a phase change is called the heat of transformation or latent heat L for the process. The total heat Q transferred in a phase change is then,
Q = Lm
Here m is the mass of the sample that changes phase.
The heat given off the steam Q_{s} will be equal to,
Q_{s} = m_{s}L_{v}+ m_{s}c_{w}ΔT
Here, mass of steam is m_{s}, latent heat vaporization is L_{v}, specific heat capacity of water is c_{w} and the temperature difference is ΔT.
The heat taken in by the ice Q_{i} will be equal to,
Q_{i} = m_{i}L_{f}+ m_{i}c_{w}ΔT
Here, mass of ice is m_{i}, latent heat fusion is L_{f}, specific heat capacity of water is c_{w} and the temperature difference is ΔT.
Heat given off the steam Q_{s} is equal to the heat taken in by the ice Q_{i}.
So, Q_{s} = Q_{i}
m_{s}L_{v}+ m_{s}c_{w}ΔT = m_{i}L_{f}+ m_{i}c_{w}ΔT
m_{s}(L_{v}+ c_{w}ΔT) = m_{i}(L_{f}+ c_{w}ΔT)
m_{s} = m_{i}(L_{f}+ c_{w}ΔT)/ (L_{v}+ c_{w}ΔT)
To obtain the mass of the steam at 100^{ °}C must be mixed with 150 g of ice at 0 ^{°}C, substitute 150 g for mass of ice m_{i}, 333×10^{3} J/kg for L_{f}, 4190 J/kg.K for c_{w}, 50^{°} C for ΔT, 2256×10^{3} J/kg for L_{v} in the equation m_{s} = m_{i}(L_{f}+ c_{w}ΔT)/ (L_{v}+ c_{w}ΔT), we get,
=(150 g)[(333×10^{3} J/kg) +(4190 J/kg.K) (50^{°} C)]/ [(2256×10^{3} J/kg)+ (4190 J/kg.K) (50^{°} C)]
=(150 g×(10^{-3} kg/1g))[(333×10^{3} J/kg) +(4190 J/kg.K) (50+273)K]/ [(2256×10^{3} J/kg)+ (4190 J/kg.K) (50+273)K]
= 0.033 kg
From the above observation we conclude that, the mass of steam at 100^{ °}C must be mixed with 150 g of ice at 0 ^{°}C would be 0.033 kg.
_____________________________________________________________________________________________________
A small electric immersion heater is used to boil 136 g of water for a cup of instant coffee. The heater is labeled 220 watts. Calculate the time required to bring this water from 23.5ºC to the boiling point, ignoring nay heat losses.
Heat Q that must be given to a body of mass m, whose material has a specific heat c, to increase its temperature from initial temperature T_{i} to final temperature T_{f} is,
Q = mc (T_{f} - T_{i})
But time (t) is equal to the heat energy (Q) divided by power (P).
t = Q/ P
= mc (T_{f} - T_{i})/P
To obtain the time required to bring this water from 23.5^{°} C to the boiling point, substitute 136 g for mass of water m, 4190 J/kg. K for specific heat capacity of water c, 100^{°} C for final temperature T_{f} (boiling point of water), 23.5° C for initial temperature T_{i} and 220 watts for power P in the equation t = mc (T_{f} - T_{i})/P,
t = mc (T_{f} - T_{i})/P
= (136 g) (4190 J/kg. K) (100^{°} C - 23.5° C) / 220 W
= (136 g×10^{-3} kg/1 g) (4190 J/kg. K) (100^{°} C - 23.5° C) / 220 W
= (0.136 kg) (4190 J/kg. K) ((100+273) K – (23.5 + 273)K) / 220 W
= (0.136 kg) (4190 J/kg. K) (373 K – 296.5 K) / 220 W
= (0.136 kg) (4190 J/kg. K) (76.5 K) / 220 W
= 198.15 s
Rounding off to three significant figures, the time required to bring this water from 23.5^{°} C to the boiling point would be 198 s.
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