Chapter 10: Thermal Properties of Matter

Ans: Given that,

Coefficient of volume expansion of glycerine, ${{\alpha }_{v}}=49\times {{10}^{-5}}{{K}^{-1}}$

Rise in temperature, T$=30{}^\circ C$ 

Fractional change in its volume = $\frac{\Delta V}{V}$ 

This change is related with the change in temperature as:

$\frac{\Delta V}{V}={{\alpha }_{V}}\Delta T$

$\Rightarrow \left( {{V}_{{{T}_{2}}}}-{{V}_{{{T}_{1}}}} \right)={{V}_{{{T}_{1}}}}{{\alpha }_{V}}\Delta T$

$\Rightarrow \frac{m}{{{\rho }_{{{T}_{2}}}}}-\frac{m}{{{\rho }_{{{T}_{1}}}}}=\frac{m}{{{\rho }_{{{T}_{2}}}}}{{\alpha }_{V}}\Delta T$ 

Where,

m = Mass of glycerine

${{\rho }_{{{T}_{1}}}}$= Initial density at ${{T}_{1}}$ 

${{\rho }_{{{T}_{2}}}}$= Final density at ${{T}_{2}}$

$\Rightarrow \frac{{{\rho }_{{{T}_{1}}}}-{{\rho }_{{{T}_{2}}}}}{{{\rho }_{{{T}_{2}}}}}={{\alpha }_{V}}\Delta T$ 

Where,

$\frac{{{\rho }_{{{T}_{1}}}}-{{\rho }_{{{T}_{2}}}}}{{{\rho }_{{{T}_{2}}}}}=$Fractional change in density

Therefore, fractional change in the density of glycerin $=49\times {{10}^{-5}}\times 30=1.47\times {{10}^{-2}}$. 

Ans: Given that,

Power of the drilling machine, $P=10kW=10\times {{10}^{3}}W$ 

Mass of the aluminum block, $m=8.0kg=8\times {{10}^{3}}g$ 

Time for which the machine is used, $t=2.5\min =2.5\times 60=150s$ 

Specific heat of aluminum, $c=0.91J{{g}^{-1}}{{K}^{-1}}$ 

Rise in the temperature of the block after drilling $=\delta T$ 

Total energy of the drilling machine = Pt

$\Rightarrow Pt=10\times {{10}^{3}}\times 150=1.5\times {{10}^{6}}J$ 

It is given that only 50% of the power is useful.

Useful energy, $\Delta Q=\frac{50}{100}\times 1.5\times {{10}^{6}}J$ 

However, $\Delta Q=mc\Delta T$

$\Rightarrow \Delta T=\frac{\Delta Q}{mc}$ 

$\Rightarrow \Delta T=\frac{7.5\times {{10}^{5}}}{8\times {{10}^{3}}\times 0.91}=103{}^\circ C$ 

Clearly, in 2.5 minutes of drilling, the rise in the temperature of the block is $103{}^\circ C$.

Ans: Given that,

Mass of the copper block, m $=2.5kg=2500g$ 

Rise in the temperature of the copper block, $\Delta \theta =500{}^\circ C$ 

Specific heat of copper, $C=J{{g}^{-1}}{}^\circ {{C}^{-1}}$ 

Heat of fusion of water, $L=335J{{g}^{-1}}$ 

The maximum heat the copper block can lose, $Q=mC\Delta \theta $ 

$\Rightarrow Q=2500\times 0.39\times 500=487500J$ 

Let ${{m}_{1}}$ be the amount of ice that melts when the copper block is placed on the ice block. 

The heat gained by the melted ice, $Q={{m}_{1}}L$ 

$\Rightarrow {{m}_{1}}=\frac{Q}{L}=\frac{487500}{335}=1455.22g$ 

Thus, the maximum amount of ice that can melt is 1.45 kg.

Ans: Given that,

Mass of the metal, m $=0.20kg=200g$

Initial temperature of the metal, ${{T}_{1}}=150{}^\circ C$ 

Final temperature of the metal, ${{T}_{2}}=40{}^\circ C$ 

Calorimeter has water equivalent of mass, $m=0.025kg=25g$

Volume of water, $V=150c{{m}^{3}}$ 

Mass (M) of water at temperature $T=27{}^\circ C:150\times 1=150g$ 

Fall in the temperature of the metal:

$\Delta T={{T}_{1}}-{{T}_{2}}=150-40=110{}^\circ C$ 

Specific heat of water, ${{C}_{W}}=4.186J/g/{}^\circ K$ 

Specific heat of the metal = C 

Heat lost by the metal, $\theta =mc\Delta T$ 

Rise in the temperature of the water and calorimeter system:

$\Delta {T}'=40-27=13{}^\circ C$ 

Heat gained by the water and calorimeter system:

$\Delta {\theta }'={{m}_{1}}{{C}_{W}}\Delta {T}'$

$\Rightarrow \Delta {\theta }'=\left( M+{m}' \right){{C}_{W}}\Delta {T}'$

Heat lost by the metal = Heat gained by the water and calorimeter system

$mc\Delta T=\left( M+{m}' \right){{C}_{W}}\Delta {T}'$

$\Rightarrow 200\times C\times 110=\left( 150+25 \right)\times 4.186\times 13$ 

$\Rightarrow C=\frac{175\times 4.186\times 13}{110\times 200}=0.43J{{g}^{-1}}{{K}^{-1}}$ 

Clearly, when some heat is lost to the surroundings, then the value of C would be smaller than the actual value.

Ans: The gases listed in the provided table are diatomic. 

Apart from the translational degree of freedom, they also possess other degrees of freedom (modes of motion). 

Heat should be supplied to increase the temperature of these gases which would further increase the average energy of all the modes of motion.

Thus, the molar specific heat of diatomic gases turns out to be more than that of monatomic gases. 

Moreover, when only the rotational mode of motion is considered, then the molar specific heat of a diatomic gas $=\frac{5}{2}R$ 

$\Rightarrow \frac{5}{2}R=\frac{5}{2}\times 1.98=4.95$cal $mo{{l}^{-1}}{{K}^{-1}}$ 

Excluding the data of chlorine, all the other observations in the given table agree with $\left( \frac{5}{2}R \right)$.

This is because at room temperature, chlorine also has vibrational modes of motion apart from rotational and translational modes of motion.

Ans: From the P-T phase diagram for $C{{O}_{2}}$ as shown in the figure above,

C is the triple point of the $C{{O}_{2}}$ phase diagram. 

This suggests that at the temperature and pressure corresponding to this point (i.e., at $56.6{}^\circ C$  and 5.11 atm), the solid, liquid, and vaporous phases of $C{{O}_{2}}$ co-exist in equilibrium.

1. What is the effect of decrease of pressure on the fusion and boiling point of $C{{O}_{2}}$?

Ans: The fusion and boiling points of $C{{O}_{2}}$ decrease as pressure decreases.

1. What are the critical temperature and pressure for $C{{O}_{2}}$ ? What is their significance?

Ans: The critical temperature and critical pressure of $C{{O}_{2}}$ are $31.1{}^\circ C$  and 73 atm respectively.

Even when there is compression to a pressure greater than 73 atm, $C{{O}_{2}}$ would not liquify above the critical temperature.

1. Is $C{{O}_{2}}$ solid, liquid or gas at a) $-70{}^\circ C$ under 1 atm, b) $60{}^\circ C$ under 10 at m, c) $15{}^\circ C$ under 56 atm?

Ans: With reference to the P-T phase diagram of $C{{O}_{2}}$ ;

a) $C{{O}_{2}}$is gaseous at $-70{}^\circ C$, under 1 atm pressure.

b) $C{{O}_{2}}$is solid at $60{}^\circ C$, under 10 atm pressure. 

c) $C{{O}_{2}}$is liquid at $15{}^\circ C$, under 56 atm pressure.

Ans: The P-T phase diagram for $C{{O}_{2}}$ is considered as shown in the question.

At 1 atm pressure and at $-60{}^\circ C$ , $C{{O}_{2}}$ positions itself to the left of $-56.6{}^\circ C$ (triple point C). Thus, it places itself in the region of vaporous and solid phases. 

Clearly, $C{{O}_{2}}$ condenses into the solid state directly without going through the liquid state.

1. What happens when $C{{O}_{2}}$ at 4 atm pressure is cooled from room temperature at constant pressure?

Ans: At 4 atm pressure, $C{{O}_{2}}$ places itself below 5.11 atm (triple point C). 

Thus, it positions itself in the region of vaporous and solid phases. 

Clearly, it condenses into the solid-state directly without passing through the liquid state.

1. Describe qualitatively the changes in a given mass of solid $C{{O}_{2}}$ at 10 atm pressure and temperature $-65{}^\circ C$as it is heated up to room temperature at constant pressure.

Ans: If the temperature of a mass of solid $C{{O}_{2}}$ (at 10 atm pressure and at $-65{}^\circ C$ ) is increased, it changes to the liquid phase and then to the vaporous phase.

It projects a line parallel to the temperature axis at 10 atm. 

Clearly, the fusion and boiling points are provided by the intersection point where this parallel line cuts the fusion and vaporization curves.

1. $C{{O}_{2}}$is heated to a temperature $70{}^\circ C$ and compressed isothermally. What changes in its properties do you expect to observe?

Ans: When $C{{O}_{2}}$ is heated to $70{}^\circ C$ and compressed isothermally, it would not show any transition to the liquid state. 

This is because $70{}^\circ C$is higher than the critical temperature of $C{{O}_{2}}$. 

Clearly, it would remain in the vapor state but would depart from its ideal behavior as pressure is increased.

Ans: Given that,

Initial temperature of the body of the child, ${{T}_{1}}=101{}^\circ F$ 

Final temperature of the body of the child, ${{T}_{2}}=98{}^\circ F$

Change in temperature, $\Delta T=\left[ \left( 101-98 \right)\times \frac{5}{9} \right]{}^\circ C$ 

Time taken to reduce the temperature, t = 20 min 

Mass of the child, $m=30kg=30\times {{10}^{3}}g$ 

Specific heat of the human body = Specific heat of water $=c=100cal/kg/{}^\circ C$ 

Latent heat of evaporation of water,  $L=580cal$ ${{g}^{-1}}$ 

The heat lost by the child is given as:

$\Delta \theta =mc\Delta T$

$\Rightarrow \Delta \theta =30\times 1000\times \left( 101-98 \right)\times \frac{5}{9}=50000cal$

Let ${{m}_{1}}$  be the mass of the water evaporated from the child’s body in 20 min. 

Loss of heat through water is given by:

$\Delta \theta ={{m}_{1}}L$

$\Rightarrow {{m}_{1}}=\frac{\Delta \theta }{L}$

$\Rightarrow {{m}_{1}}=\frac{50000}{580}=86.2g$

Therefore, average rate of extra evaporation caused by the drug $=\frac{{{m}_{1}}}{t}$ 

$\Rightarrow \frac{{{m}_{1}}}{t}=~~\frac{86.2}{200}=4.3g/\min $

Ans: Given that,

Side of the given cubical ice box, $s=30cm=0.3m$ 

Thickness of the ice box, $l=5.0cm=0.05m$ 

Mass of ice kept in the ice box, m = 4 kg

Time gap, $t=6h=6\times 60\times 60s$

Outside temperature, $T=45{}^\circ C$ 

Coefficient of thermal conductivity of thermacole, $K=0.01J{{s}^{-1}}{{m}^{-1}}{{K}^{-1}}$   

Heat of fusion of water,  $L=335\times {{10}^{3}}Jk{{g}^{-1}}$ 

Assume ${m}'$ be the total amount of ice that melts in 6 h. 

The amount of heat lost by the food:

$\theta =\frac{KA\left( T-0 \right)t}{l}$ 

where,

$A=$ Surface of the box $=6{{s}^{2}}=6\times {{\left( 0.3 \right)}^{2}}=0.54{{m}^{3}}$ 

$\Rightarrow \theta =\frac{0.01\times 0.54\times 45\times 6\times 60\times 60}{0.05}=104976J$ 

However, $\theta ={m}'L$ 

$\Rightarrow {m}'=\frac{\theta }{L}$

$\Rightarrow {m}'=\frac{104976}{335\times {{10}^{3}}}=0.313g$

Mass of ice left $4-0.313=3.687kg$ 

Clearly, the amount of ice remaining after 6h is 3.687kg.

Ans: Given that,

Base area of the boiler, $A=0.15{{m}^{2}}$ 

Thickness of the boiler, \[l=1.0\text{ }cm=0.01\text{ }m\] 

Boiling rate of water, R = $6.0kg/\min $ 

Mass, m = 6 kg 

Time, t = 1 min = 60 s 

Thermal conductivity of brass, K= $=109J{{s}^{-1}}{{m}^{-1}}{{K}^{-1}}$ 

Heat of vaporization, L$=2256\times {{10}^{3}}Jk{{g}^{-1}}.$

The amount of heat flowing into water through the brass base of the boiler is given by:

$\theta =\frac{KA\left( {{T}_{1}}-{{T}_{2}} \right)t}{l}$

Where, 

${{T}_{1}}$ = Temperature of the flame in contact with the boiler 

${{T}_{2}}$ = Boiling point of water $=100{}^\circ C$ 

Heat required for boiling the water: 

$\theta =mL$ 

Equating equations (a) and (b), we get:

$mL=\frac{KA\left( {{T}_{1}}-{{T}_{2}} \right)t}{l}$ 

$\Rightarrow \left( {{T}_{1}}-{{T}_{2}} \right)=\frac{mLl}{KAt}$ 

$\Rightarrow \left( {{T}_{1}}-{{T}_{2}} \right)=\frac{6\times 2256\times {{10}^{3}}\times 0.001}{109\times 0.15\times 60}=137.98{}^\circ C$

Clearly, the temperature of the part of the flame in contact with the boiler is $137.98{}^\circ C$.