Thermal Properties of Matter

11.13 Mass of the copper block, m = 2.5 kg = 2.5 $*{10}^3$ gm

Rise in temperature of the copper block,  $?$ T = 500°C

Specific heat of the copper, C = 0.39 g–1 K–1

Heat of fusion of water, L = 335 J g–1

The maximum heat the copper block can lose, Q = mc $?$ T = 2.5 $*{10}^3*0.39*500$ = 487500 J

Let $m_1$ gm be the mass of the ice, which will melt because of the copper block.

Heat gained by ice block = Q = $m_{1}L$

$m_1=\frac{Q}{L}$ = $\frac{487500}{335}$ g = 1455.22 gm = 1.45 kg

11.11 Coefficient of volume expansion of glycerin, ${\alpha }_V$ = 49 $*{10}^{-5}$ /K

Rise in temperature, $?T$ = 30°C

Fractional change in volume = $\frac{?V}{V}$

We can write, $\frac{?V}{V}$ = ${\alpha }_V*?T$ = 49 $*{10}^{-5}*30$ = 0.0147 ……(i)

If the final volume is $V_2$ and initial volume is $V_1$ , then

$\frac{?V}{V}$ = $\frac{V_2-V_1}{V_1}$

$V_2=\frac{m}{{\rho }_2}$ and $V_1=\frac{m}{{\rho }_1}$ where ${\rho }_1$ & ${\rho }_2$ are initial and final densities

$\frac{?V}{V}$ = $\frac{V_2-V_1}{V_1}$ = $\frac{{\rho }_2-{\rho }_1}{{\rho }_1}$ = fractional change in density = 0.0147 = 1.47 $*$ ${10}^{-2}$

11.10 Initial temperature, $T_1$ = 40.0°C, Final temperature, $T_2$ = 250°C, $?$ T = $T_2$ - $T_1$ = 210°C

Initial length of the brass rod at $T_1$ , $l_b$ = 50 cm, Initial diameter of the brass rod at $T_1$ , $d_1$ = 3 mm

Length of the steel rod $l_s=50cm$

For the expansion of the brass rod, we have:

$\frac{Changeinlength(?l_{b)}}{Originallength(l_{b)}}$ = ${\alpha }_b?T$ , then $?l_b$ = 50 $*2.0\mathrm{}*\mathrm{}{10}^{-5}*210$ = 0.21 cm

For the expansion of the steel rod, we have:

$\frac{Changeinlength(?l_{s)}}{Originallength\left(l_s\right)}$ = ${\alpha }_b?T$ , then $?l_s$ = 50 $*1.2\mathrm{}*\mathrm{}{10}^{-5}*210$ = 0.126 cm

Total change in length = 0.21 + 0.126 = 0.336 cm

Since the rods are free at the end, no thermal stress developed.

11.9 Initial temperature, $T_1$ = 27°C, Length of the wire $l_1$ at $T_1$ = 1.8 m

Final temperature, $T_2$ = -39°C

Diameter of the wire, d = 2.0 mm = 2 $*{10}^{-3}$ m

Coefficient of linear expansion of brass, $\alpha =2*{10}^{-5}$ /K

Youngs' modulus of brass, Y = 0.91 $*{10}^{11}$ Pa

Let the tension developed be F

We know Youngs' modulus = $\frac{Stress}{Strain}$ = $\frac{\frac{F}{A}}{\frac{?L}{L}}=Y$

Y $*\frac{?L}{L}$ = $\frac{F}{A}$ or F = $\frac{AY?L}{L}$

Here, A = cross-sectional area of the wire = $\frac{\pi }{4}{d}^2$ = $\frac{\pi }{4}{(2\mathrm{}*{10}^{-3})}^2$ $m^2$ = 3.1416 $*{10}^{-6}{m}^2$

Now $?L$ can be written as $?L$ = $\alpha L(T_2-\mathrm{}{T}_1)$ = ( $2*{10}^{-5})*L*(-39-27)$ = -1.32 $*{10}^{-3}$ L

Substituting all values, we get

F = (3.1416 $*{10}^{-6}*0.91\mathrm{}*{10}^{11}*(-1.32*{10}^{-3})$ = -377.37 N

11.8 Initial temperature, $T_1$ = 27.0 °C, Initial diameter of the hole, $d_1$ = 4.24 cm

Final temperature, $T_2$ = 227.0 °C, Final diameter of the hole $=d_2$

Coefficient of linear expansion of copper, ${\alpha }_{copper}$ = 1.70 $*{10}^{-5}$ K–1

We know

$\frac{Changeinarea(?A)}{Originalarea\left(A\right)}$ = $\beta ?$ T where $\beta$ is the coefficient of superficial expansion, $\beta =2{\alpha }_{copper}$

$\frac{(\mathit{\pi }\frac{{\mathit{d}}_2^2}{4}-\mathit{}\mathit{\pi }\frac{{\mathit{d}}_1^2}{4})}{\mathit{\pi }\frac{{\mathit{d}}_1^2}{4}}$ = $2{\alpha }_{copper}?$ T

$\frac{{\mathit{d}}_2^2-\mathit{}{\mathit{d}}_1^2}{{\mathit{d}}_1^2}$ = $2{\alpha }_{copper}?$ T

$\frac{{\mathit{d}}_2^2}{{\mathit{d}}_1^2}$ - 1= $2{\alpha }_{copper}\left(T_2-T\right)=$ 2 $*$ 1.70 $*{10}^{-5}$ (227-27) = 6.8 $*{10}^{-3}$

$d_2^2=1.0068*4.24*4.24$

$d_2$ = 4.2544 cm

So change in diameter = 4.2544 – 4.24 = 0.01439 cm

Diameter increase by 1.44 $*{10}^{-2}$ cm

11.7 Given, temperature $T_1$ = 27 °C = 27 + 273.16 K = 300.16 K

Outer dia of the shaft at temp $T_1$ ,  $d_1$ = 8.7 cm

Diameter of the central hole of the wheel,  $d_2$ = 8.69 cm

The change in diameter, Δd= $d_2-d_1=$ 8.69 – 8.7 = $-$ 0.01 cm

After the shaft is cooled in dry ice, its temperature becomes $T_2$ . It can be calculated from the relation

Δd= $d_1*{\alpha }_{steel}$  ( $T_2-T_1)$

-0.01 = 8.7 $*1.20\mathrm{}*\mathrm{}{10}^{-5}(T_2-300)$

$T_2-300$ = -95.78

$T_2$ = 204.22 K = -68.94 $?$$$

11.6 Length of the steel tape, l = 1 m = 100 cm, At temperature T = 27 $°$ C

Coefficient of linear expansion of steel $\alpha$ = 1.2 $*{10}^{-5}$ / K

Let $l_1=63cm$ be the length of the steel rod at temperature $T_1$ = 45.0 °C and

$l_2$ be the length of the steel rod and l' be the length of the steel tape at 45.0 °C

We have l' = l + $\alpha l(T_1-T)$ = 100 + 1.2 $*{10}^{-5}*100*$  (45-27) = 100.0216 cm

$l_2$ can be calculated as $l_2$ = $\frac{l\text{'}}{l}*63$ = 63.0136 cm