Solved Examples on Heat Phenomena:-

Question 1:-

(a) Calculate the rate of heat loss through a glass window of area 1.4 m² 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.

H = A(T₂-T₁)/∑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.

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 T₂ is the inside temperature and T₁ is the outside temperature, then the temperature difference ΔT in ^°C will be,

ΔT = T₂-T₁

     = 5^°C /9 ^°F (72^°F –(-20 ^° F))

     = 51.1  ^°C

     = (51.1+273) K

     = 324.1 K

 H = kA ΔT/ Δx

     = (1.0 W/m.K) (1.4 m²) (324.1 K)/ (3.0 mm) 

     = (1.0 W/m.K) (1.4 m²) (324.1 K)/ (3.0 mm×10^-3 m/1 mm)

    = 2.4×10⁴ W

From the above observation we conclude that, the rate of heat loss through a glass window would be 2.4×10⁴ W.

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 m².K/W

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 m².K/W

H = A ΔT/ (R_g+ R_a)

    = (1.4 m²) (324.1 K) /(3.0×10^-3 m².K/W) (2.88 m².K/W)

    = 25 W

From the above observation we conclude that, the corresponding rate of heat loss H would be 25 W.

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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.

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_sL_v+ m_sc_wΔT

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_iL_f+ m_ic_wΔT

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_sL_v+ m_sc_wΔT = m_iL_f+ m_ic_wΔT

m_s(L_v+ c_wΔT) = m_i(L_f+ c_wΔT)

m_s = m_i(L_f+ c_wΔT)/ (L_v+ c_wΔT)

m_s = m_i(L_f+ c_wΔT)/ (L_v+ c_wΔT)

=(150 g)[(333×10³ J/kg) +(4190 J/kg.K) (50^° C)]/ [(2256×10³ J/kg)+ (4190 J/kg.K) (50^° C)]

 =(150 g×(10^-3 kg/1g))[(333×10³ J/kg) +(4190 J/kg.K) (50+273)K]/ [(2256×10³ 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.

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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

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.