**Solved Problems on Specific Heat, Latent Heat and Entropy:- **

**Problem 1:-**

A spherical constant temperature heat source of radius r_{1} is at the center of a uniform solid sphere of radius r_{2}. Find out the rate which is proportional to heat transferred through the surface of the sphere.

**Solution:-**

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*

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

From above we know that, the rate *H* at which heat is transferred through the slab is directly proportional to the area (*A*) available.

Area *A* of solid sphere is defined as,

*A* = 4π*r*^{2}

Here *r* is the radius of sphere.

So, the area *A*_{1} of uniform small solid sphere having radius *r*_{1} will be,

*A*_{1} = 4π*r*_{1}^{2}

And, the area *A*_{2} of uniform large solid sphere having radius *r*_{2} will be,

*A*_{2} = 4π*r*_{2}^{2}

Thus the area *A* from which heat is transferred through the surface of the sphere will be the difference of area of uniform large solid sphere* A*_{2} and small solid sphere* A*_{1}.

So, *A* = *A*_{2} - *A*_{1}

= 4π*r*_{2}^{2} - 4π*r*_{1}^{2}

= 4π (*r*_{2}^{2} - *r*_{1}^{2})

Since the rate *H* at which heat is transferred through the slab is directly proportional to the area (*A*) available, therefore the rate at which heat is transferred through the surface of the sphere is proportional to *r*_{2}^{2} - *r*_{1}^{2}.

**Problem 2:-**

What mass of steam at 100°C must be mixed with 150 g of ice at 0°C, in a thermally insulated container, to produce liquid water at 50°C.

**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*/*m*Δ*T*

So, *Q* = *c m*Δ*T*

Here, the heat transferred is *Q*, specific heat capacity is *c*, mass is *m* and the temperature difference is Δ*T*.

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}Δ*T*

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 Δ*T*.

The heat taken in by the ice *Q*_{i} will be equal to,

*Q*_{i} = *m*_{i}*L*_{f}+* m*_{i}*c*_{w}Δ*T*

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 Δ*T*.

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}Δ*T* = *m*_{i}*L*_{f}+* m*_{i}*c*_{w}Δ*T*

*m*_{s}(*L*_{v}+* c*_{w}Δ*T*) = *m*_{i}(*L*_{f}+* c*_{w}Δ*T*)

*m*_{s} = *m*_{i}(*L*_{f}+* c*_{w}Δ*T*)/ (*L*_{v}+* c*_{w}Δ*T*)

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 Δ*T*, 2256×10^{3} J/kg for *L*_{v} in the equation *m*_{s} = *m*_{i}(*L*_{f}+* c*_{w}Δ*T*)/ (*L*_{v}+* c*_{w}Δ*T*), we get,

*m*_{s} = *m*_{i}(*L*_{f}+* c*_{w}Δ*T*)/ (*L*_{v}+* c*_{w}Δ*T*)

=(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.

**Problem 3:-**

(a) Compute the possible increase in temperature for water going over Niagara Falls, 49.4 m high. (b) What factors would tend to prevent this possible rise?

**Concept:-**

Work done *W* is defined as,

*W* = *mg*Δ*y*

Here *m* is the mass, *g* is the free fall acceleration and Δ*y* is the increase in height.

The specific heat capacity *c* of a material is equal to the heat capacity *C* per unit mass *m* of the body.

So, *c* = *C*/*m*

= *Q*/*m*Δ*T* (Since, *C* = *Q*/Δ*T* )

Here Δ*T* is the increase in temperature.

So, *Q* = *mc*Δ*T*

As, |*Q|* = |*W|*,

*mc*Δ*T* = *mg*Δ*y*

So, Δ*T* = *g*Δ*y*/*c*

**Solution:-**

(a) To obtain the possible increase in temperature Δ*T*, substitute 9.81 m/s^{2} for *g*, 49.4 m for Δ*y* and 4190 J/kg.K for specific heat capacity of water *c* in the equation Δ*T* = *g*Δ*y*/*c*,

Δ*T* = *g*Δ*y*/*c*

= (9.81 m/s^{2}) (49.4 m)/ (4190 J/kg.K)

= (9.81 m/s^{2}) (49.4 m)/ (4190 J/kg.K) (1 kg.m^{2}/s^{2} /1 J)

= 0.116 K

From the above observation we conclude that, the possible increase in temperature Δ*T* would be 0.116 K

**Problem 4:-**

In a certain solar house, energy from the Sun is stored in barrels filled with water. In a particular winter stretch of five cloudy days, 5.22 GJ are needed to maintain the inside of the house at 22.0°C. Assuming that the water in the barrels is at 50.0°C, what volume of water is required?

**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})

So, *m* = *Q/ c *(*T*_{f} - *T*_{i})

Density *ρ* is equal to mass *m* per unit volume *V*.

So, *ρ* = *m*/*V*

So volume *V* will be,

*V* = *m*/*ρ*

**Solution:-**

To find the volume water, first we have to find out the mass of water which is required to transfer 5.22 GJ amount of heat energy.

To find the mass *m* of water, substitute 5.22 GJ for *Q*, 4190 J/kg. K for specific heat capacity *c* of water, 50.0 ^{°} C for *T*_{f} and 22.0 ° C for *T*_{i} in the equation *m* = *Q/ c *(*T*_{f} - *T*_{i}),

*m* = *Q/ c *(*T*_{f} - *T*_{i})

= 5.22 GJ/(4190 J/kg. K) (50.0 ^{°} C-22.0 ° C)

= (5.22 GJ) (10^{9} J/1 GJ)/(4190 J/kg. K) ((50.0+273) K –(22.0+273) K)

= (5.22 ×10^{9} J)/(4190 J/kg. K) (28 K)

= 4.45×10^{4} kg

To obtain the volume *V* of water, substitute 4.45×10^{4} kg for mass *m* and 998 kg/m^{3} for density* ρ* of water in the equation *V* = *m*/*ρ*,

*V* = *m*/*ρ*

= (4.45×10^{4} kg) / (998 kg/m^{3})

= 44.5 m^{3}

From the above observation we conclude that, the volume *V* of water will be 44.5 m^{3}.

**Problem 5:-**

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 23.5°C to the boiling point, ignoring any 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.

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