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

Hat 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 Δ

x.(c) directly proportional to the temperature difference Δ

T.So,

H=kAΔT/ ΔxHere

kis the proportionality constant and is called thermal conductivity of the rod.Again the rate of heat transfer

His defined as,

H=A(T_{2}-T_{1})/∑R_{n}=

AΔT/ (R_{g}+R_{a})Here

Ais the area, ΔTis the temperature difference,R_{g}is the thermal resistance of glass at the thickness 3.0 mm andR_{a}is the thermal resistance of air at the thickness 7.5 cm.The thermal resistance or

R-value, defined by,

R=L/kHere

L is the thickness of the material through which the heat is transferred andkis 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 andT_{1}is the outside temperature, then the temperature difference ΔTin^{°}C will be,Δ

T=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 ΔTand 3.0 mm for Δxin the equationH=kAΔT/ Δx, we get,

H=kAΔT/ 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}WFrom 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 resistanceR_{g}of galss at the thickness 3.0 mm andR_{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 forLand 1.0 W/m.K (thermal conductivity of window glass) forkin the equationR=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/WTo obtain the thermal resistance

R_{a}of air at the thickness 7.5 cm, substitute 7.5 cm forLand 0.026 W/m.K (thermal conductivity of dry air) forkin the equationR=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/WTo find out the corresponding rate of heat loss

H, substitute 1.4 m^{2}forA,3.0×10^{-3}m^{2}.K/W forR_{g}and 2.88 m^{2}.K/W forR_{a}in the equationH=AΔT/ (R_{g}+R_{a}), we get,

H=AΔT/ (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

Hwould 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/mΔTSo,

Q=c mΔTHere, the heat transferred is

Q, specific heat capacity isc, mass ismand 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

Lfor the process. The total heatQtransferred in a phase change is then,

Q=LmHere

mis 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}ΔTHere, mass of steam is

m_{s}, latent heat vaporization isL_{v}, specific heat capacity of water isc_{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}ΔTHere, mass of ice is

m_{i}, latent heat fusion isL_{f}, specific heat capacity of water isc_{w}and the temperature difference is ΔT.Heat given off the steam

Q_{s}is equal to the heat taken in by the iceQ_{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 icem_{i}, 333×10^{3}J/kg forL_{f}, 4190 J/kg.K forc_{w}, 50^{°}C for ΔT, 2256×10^{3}J/kg forL_{v}in the equationm_{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

Qthat must be given to a body of massm, whose material has a specific heatc, to increase its temperature from initial temperatureT_{i}to final temperatureT_{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 waterm, 4190 J/kg. K for specific heat capacity of waterc, 100^{°}C for final temperatureT_{f}(boiling point of water), 23.5° C for initial temperatureT_{i}and 220 watts for powerPin the equationt=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.