**Solved Examples on Heat Transfer:-**

**Question 1:-**

An idealized representation of the air temperature as a function of distance from a single-pane window on a calm, winter day is shown in below figure. The window dimensions are 60 cm × 60 cm × 0.50 cm. *(a)* At what rate does heat flow out through the window? (Hint: The temperature drop across the glass is very small.) *(b)* Estimate the difference in temperature between the inner and outer glass surfaces.

**Concept**:-

The rate *H* at which heat is transferred through the slab is,

(a) directly proportional to the area (*A*) available.

(b) inversely proportional to the thickness of the slab Δ*x*.

(c) directly proportional to the temperature difference Δ*T*.

So, *H* = *kA* Δ*T*/ Δ*x*

Here, *k* is the proportionality constant and is called thermal conductivity of the material.

From the equation *H* = *kA* Δ*T*/ Δ*x*, the temperature difference Δ*T* will be,

Δ*T* = *H* Δ*x*/*kA*

The temperature gradient is defined as,

Temperature gradient = Δ*T*/ Δ*x*

**Solution**:-

(a)To find out the rate of heat flows out through the window, first we have to find out the temperature gradient both inside and outside the window.

Inside, the temperature gradient Δ*T*/ Δ*x* will be,

Δ*T*/ Δ*x* = (20^{°} C- 5^{°} C)/8 cm (Since, Δ*x* = 8 cm)

= ((20 +273)K- (5+273) K)/(8 cm×10^{-2} m/ 1 cm)

= 293 K-278 K/0.08 m

= 187.5 K/m

Rounding off to two significant figures the temperature gradient will be 190 K/m.

Similar result will be occur for also outside.

Area *A* from which heat will flow from the window will be,

*A* = (60 cm ×60 cm)^{2}

= [(60 cm×10^{-2} m/ 1 cm) (60 cm×10^{-2} m/ 1 cm)]^{2}

= (0.6 m)^{2}

To find out the heat flow *H* through the air, substitute 0.026 W/m. K for thermal conductivity *k*, (0.6 m)^{2} for *A* and 190 ^{°} C/m for Δ*T*/ Δ*x* in the equation *H* = *kA* Δ*T*/ Δ*x*,

*H* = *kA* Δ*T*/ Δ*x*

= (0.026 W/m. K) (0.6 m)^{2} (190 K/m)

= 1.8 W

The value that we arrived at is the rate that heat flows through the air across an area the size of the window on either side of the window. Therefore the rate of heat flows out through the window would be 1.8 W.

(b) To find out the difference in temperature Δ*T* between the inner and outer glass surfaces, substitute 1.8 W for *H*, 0.50 cm for Δ*x*, 1.0 W/m. K for *k* and (0.6 m)^{2} for *A* in the equation Δ*T* = *H* Δ*x*/*kA*,

Δ*T* = *H* Δ*x */ *kA*

= (1.8 W) (0.50 cm)/( 1.0 W/m. K)( (0.6 m)^{2})

= (1.8 W) (0.50 cm×10^{-2} m/1 cm )/( 1.0 W/m. K)( (0.6 m)^{2})

= 0.025 K

From the above observation we conclude that, the difference in temperature Δ*T* between the inner and outer glass surfaces would be 0.025 K.

________________________________________________________________________________________

**Question 2:-**

Two identical rectangular rods of metal are welded end to end as shown in below figure *(a)*, and 10 J of heat flows through the rods in 2.0 min. How long would it take for 30 J to flow through the rods if they are welded as shown in below figure *(b)*.

**Concept:-**

The rate *H* at which heat is transferred through the slab is,

(a) directly proportional to the area (*A*) available.

(b) inversely proportional to the thickness of the slab Δ*x*.

(c) directly proportional to the temperature difference Δ*T*.

So, *H* = *kA* Δ*T*/ Δ*x*

Here *k* is the proportionality constant and is called thermal conductivity of the material.

**Solution:-**

As the temperature difference in this case Δ*T* = *T*_{hot} – *T*_{cold} is same for both the case, thus the thermal conductivity will be same for both the cases.

To obtain the rate of heat transfer *H*_{o} for the rectangular rod which are connected in series (configuration (*a*)), substitute 2*L *for Δ*x* in the equation *H*_{o} = *kA* Δ*T*/ Δ*x*,

*H*_{o} = *kA* Δ*T*/ Δ*x*

= *kA* Δ*T*/2*L*

To obtain the rate of heat transfer (*H*_{o})^{'} for the rectangular rod which are connected in parallel (configuration (*b*)), substitute 2*A *for *A* in the equation *H*_{o} = *kA* Δ*T*/ Δ*x*,

(*H*_{o})^{'} = *kA* Δ*T*/ Δ*x*

= *k *(2*A*) Δ*T*/*L*

So, (*H*_{o})^{'} /* H*_{o} = (*k *(2*A*) Δ*T*/*L)*/ (*kA* Δ*T*/2*L*)

= 4

So, (*H*_{o})^{'} = 4* H*_{o}

Again also rate of heat transfer* H* is defined as,

*H* = *W*/*t*

Here *W* is the energy and *t* is the time.

So, time *t* will be,

*t =* *W*/*H*

To find the rate of heat transfer *H*_{o} for the rectangular rod which are connected in series (configuration (*a*)), substitute 10 J for *W* and 2 min for *t *in the equation* t =* *W*/*H*,

*H*_{o} = *W*/*t*

= 10 J/ 2 min

= 5 J/min

To find out (*H*_{o})^{'}, substitute 5 J/min for *H*_{o} in the equation (*H*_{o})^{'} = 4* H*_{o},

(*H*_{o})^{'} = 4* H*_{o}

= 4(5 J/min)

= 20 J/min

Using equation* t =* *W*/*H*, the time *t* for 30 J to flow through the rods if they were welded in the figure parallel configuration will be,

*t* = *W*/(*H*_{o})^{'}

To obtain the time *t* for 30 J to flow through the rods if they were welded in the figure parallel configuration, substitute 30 J for *W* and 20 J/min for (*H*_{o})^{'} in the equation *t* = *W*/(*H*_{o})^{'},

*t* =*W*/(*H*_{o})^{'}

=(30 J)/(20 J/min)

= 1.5 min

From the above observation we conclude that, the time *t* for 30 J to flow through the rods if they were welded in the figure parallel configuration would be 1.5 min.

_______________________________________________________________________________________________

**Question 3:-**

Ice has formed on a shallow pond and a steady state has been reached with the air above the ice at -5.2ºC and the bottom of the pond at 3.98ºC. If the total depth of ice + water is 1.42 m, how thick is the ice? (Assume that the thermal conductivities of ice and water are 1.67 and 0.502 W/m.K, respectively.)

**Concept**:-

The temperature *T*_{x} at the interface of a compound slab is equal to (*T*_{1}*R*_{2} +* T*_{2} *R*_{1}) /(*R*_{1} + *R*_{2}).

So, *T*_{x} = (*T*_{1}*R*_{2} +* T*_{2} *R*_{1}) /(*R*_{1} + *R*_{2})

Here, *T*_{1}, *T*_{2} are the temperature of two surfaces (with T_{2} > T_{1}) and *R*_{1},* R*_{2} are thermal resistance of the two materials.

**Solution**:-

At the interface between ice and water *T*_{x} = 0^{°} C

Substitute 0^{°} C for *T*_{x} in the equation *T*_{x} = (*T*_{1}*R*_{2} +* T*_{2} *R*_{1}) /(*R*_{1} + *R*_{2}),

(*T*_{1}*R*_{2} +* T*_{2} *R*_{1}) /(*R*_{1} + *R*_{2}) = *T*_{x}

(*T*_{1}*R*_{2} +* T*_{2} *R*_{1}) /(*R*_{1} + *R*_{2}) = 0

So, *T*_{1}*R*_{2} +* T*_{2} *R*_{1} = 0

Or, *k*_{1}*T*_{1}/*L*_{1} + *k*_{2}*T*_{2}/*L*_{2} = 0

(*k*_{1}*T*_{1}*L*_{2} + *k*_{2}*T*_{2}* L*_{1}) /* L*_{1}* L*_{2} = 0

*k*_{1}*T*_{1}*L*_{2} + *k*_{2}*T*_{2}* L*_{1} = 0

*k*_{1}*T*_{1}*L*_{2} + *k*_{2}*T*_{2} (*L*-*L*_{2}) = 0

*k*_{1}*T*_{1}*L*_{2}+ *k*_{2}*T*_{2} *L*-* k*_{2}*T*_{2}* L*_{2} = 0

*k*_{2}*T*_{2} *L* = *k*_{2}*T*_{2}* L*_{2} -* k*_{1}*T*_{1}*L*_{2}

= (*k*_{2}*T*_{2}- *k*_{1}*T*_{1})* L*_{2}

Or, *L*_{2} = *k*_{2}*T*_{2} *L*/(*k*_{2}*T*_{2}- *k*_{1}*T*_{1})

Here *L* is the depth of ice + water and *L*_{2} is the depth of ice.

To find out the thickness* L*_{2} of ice (depth of ice), substitute 1.67 W/m. K for thermal conductivity of ice *k*_{2}, -5.20^{°} C for *T*_{2}, 1.4 m for *L*, 0.502 W/m . K for thermal conductivity of water *k*_{1} and 3.98^{°} C for* T*_{1} in the equation *L*_{2} = *k*_{2}*T*_{2} *L*/(*k*_{2}*T*_{2}- *k*_{1}*T*_{1}),

*L*_{2} = *k*_{2}*T*_{2} *L*/(*k*_{2}*T*_{2}- *k*_{1}*T*_{1})

=(1.67 W/m. K) (-5.20^{°} C) (1.4 m) / ((1.67 W/m. K) (-5.20^{°} C) –(0.502 W/m . K) (3.98^{°} C))

=(1.67 W/m. K) (-5.20 +273) K (1.4 m) / ((1.67 W/m. K) (-5.20 + 273) K–(0.502 W/m . K) (3.98 +273) K)

= 635.06 W/(447.23 -139.04) W/m

= 2.06 m

Rounding off to one significant figure, the thickness of the ice will be 2 m.