physics-class-11-unit-11-chapter-03-thermal-properties-of-matter-3-lecture-3-4-mlx2u8tu04g

### <Success style="font-size:25px"> Thermal Properties Of Matter 3 L-3 </Success>
### <primary style="font-size:24px"> Latent Heat </primary>
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- Change of stale : Latent heat
	
- Latent heat: It is heat per kg. that must be added or removed when a substance undregoes a change in phase from one to another.
	
- Latent heat is accompanied with no change in temperature.

- SI unit is J/kg.

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<footer style = "font-size: 18px"> $\rightarrow$ Thermal Properties of Matter $\rightarrow$ <span style = "color:blue"> Latent Heat </span> $\rightarrow$ Latent Heat $\rightarrow$ Thermal Properties of Matter </footer>

### <Success style="font-size:25px"> Thermal Properties Of Matter 3 L-3 </Success>
### <primary style="font-size:24px"> Latent Heat </primary>
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- Three kinds of Latent heat:

- 1. $L_M$ : Latent heat of melting / fusion.

- 2. $L_V$ : Latent heat of vaporisation

- 3. $L_S$ : Latent heat of sublimation

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<footer style = "font-size: 18px">Thermal Properties of Matter $\rightarrow$ Latent Heat $\rightarrow$ <span style = "color:blue"> Latent Heat </span> $\rightarrow$ Thermal Properties of Matter $\rightarrow$ Heat Transfer </footer>

### <Success style="font-size:25px"> Thermal Properties Of Matter 3 L-3 </Success>
### <primary style="font-size:24px"> Thermal Properties of Matter </primary>
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| Material | Melting point$(T_M)$$(^\circ C)$  |$L_M$$(J/kg)$ | Boiling point$(T_B)$$(^\circ C)$ | $L _v$ $ (J/kg)$ |
| -------- | -------- | -------- | -------- | -------- |
| Ammonia    |  -77.8    | 33.2$ \times10^4$   | -33.4  | 13.7$\times10^5$ |
| Benzene  | 5.5    | $12.6\times10^4 $ | 80.1  | $3.94\times10^5 $ |
| Mercury   | -38.9| $1.14\times10^4 $| 356.6   | $2.96\times10^5$  |
| Water   | 0.0     |$33.5\times10^4 $| 100   | $22.6\times10^5  $|

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<footer style = "font-size: 18px">Latent Heat $\rightarrow$ Latent Heat $\rightarrow$ <span style = "color:blue"> Thermal Properties of Matter </span> $\rightarrow$ Heat Transfer $\rightarrow$ Convection </footer>

### <Success style="font-size:25px"> Thermal Properties Of Matter 3 L-3 </Success>
### <primary style="font-size:24px"> Heat Transfer </primary>
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- (i) Convection

- (ii) Conduction

- (iii) Radiation
	
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<footer style = "font-size: 18px">Latent Heat $\rightarrow$ Thermal Properties of Matter $\rightarrow$ <span style = "color:blue"> Heat Transfer </span> $\rightarrow$ Convection $\rightarrow$ Conduction </footer>

### <Success style="font-size:25px"> Thermal Properties Of Matter 3 L-3 </Success>
### <primary style="font-size:24px"> Conduction </primary>
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- Conduction is a process where heat is transfer directly through a material.

- $Q  \propto A $
 
 - $ \propto(T_1-T_2) $
 
 - $\propto t $
 
 - $\propto \frac{1}{L}$
 
 - $Q=K \frac{A \Delta T t}{L}$
 
-  K : Thermal conductivity $(Wm^{-1}K{-1})$

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<footer style = "font-size: 18px">Heat Transfer $\rightarrow$ Convection $\rightarrow$ <span style = "color:blue"> Conduction </span> $\rightarrow$ Problem $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Thermal Properties Of Matter 3 L-3 </Success>
### <primary style="font-size:24px"> Problem </primary>
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- An iron bar $(l_1=0.1{m}, A_1=0.02{m}^2, K_1=79 {Wm}^{-1} k^{-1})$ and brass bar $(l_2=0.1 {m}, k_2=0.02 {m}^2, k_2=109{wm}^{-1} k^{-1})$ are soldered end-to-end. The free ends of the iron bar $T_2$ and the brass bar are kept at $T_1=373 {K}$ and $T_2=273 {K}$respectively.

- 1. Obtain the temperature of the junction,$T_{0}\cdot $
- 2. The equivalent thermal conductivity of the compound bar.
- 3. The heat current through the compound bar.

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<footer style = "font-size: 18px">Convection $\rightarrow$ Conduction $\rightarrow$ <span style = "color:blue"> Problem </span> $\rightarrow$ Solution $\rightarrow$ Radiation </footer>

### <Success style="font-size:25px"> Thermal Properties Of Matter 3 L-3 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- $H_1=\frac{k_1 A_1\left(T_1-T_2\right)}{4}=\frac{k_2 A_2\left(T_1-T_2\right)}{L_2}= H_2$

- $A_1  =A_2$

- $ l_1  =l_2$

- $k_1(T_1-T_0)=k_2(T_1-T_2) \Rightarrow T_0=\frac{k_1 T_1+k_2 T_2}{k_1+k_2} .$

- $H=H_1=H_2=\frac{k_1 A(T_1-T_0)}{L}=\frac{k_2 A(T_0-T_2)}{-4} $

- $\frac{A(T_1-T_2)^L}{L(\frac{1}{K_1}+K_2)}=\frac{k^{\prime}(T_1-T_2)^2}{L} \quad k^{\prime}=\frac{2 k_1 k_2}{k_1+k_2} $

- $H^{\prime}=H=\frac{k^{\prime} A(T_1-T_2)}{2 L}=(\frac{2 k_1 k_2 A}{k_1+k_2}) \frac{A(T_1-T_2)}{2 L}$

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<footer style = "font-size: 18px">Conduction $\rightarrow$ Problem $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Radiation $\rightarrow$ Stefan–Boltzmann Law of Radiation </footer>

### <Success style="font-size:25px"> Thermal Properties Of Matter 3 L-3 </Success>
### <primary style="font-size:24px"> Stefan–Boltzmann Law of Radiation </primary>
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- The radiant energy which is the thermal radiation $Q$ emitted in a time $t$ by an object that has an absolute temperature $T$, a surface area $A$ and emissivity, $e$ is given by,

- $Q=e \sigma T^4 A t$

- Stefan–Boltzmann constant $\sigma $ = $=5.67 \times{10}^{-8}{J} /(S_ {m^2}-k^4)$

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<footer style = "font-size: 18px">Solution $\rightarrow$ Radiation $\rightarrow$ <span style = "color:blue"> Stefan–Boltzmann Law of Radiation </span> $\rightarrow$ Newton's law of cooling $\rightarrow$ Thank You </footer>

### <Success style="font-size:25px"> Thermal Properties Of Matter 3 L-3 </Success>
### <primary style="font-size:24px"> Newton's law of cooling </primary>
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- $\frac{d Q}{d t} \propto \Delta T $

- $\frac{d Q}{d t}=K \Delta T$

- Temperature of body = $T_2$

- Temperature of surrounding =$T_1$

- In time interval dt, the temperature of the body falls by $dT_2$.

- $\frac{d Q}{d t}=m c \frac{d T_2}{d t}=-K\left(T_2-T_1\right)$

- $ -m c \frac{d T_2}{d t}=k(T_2-T_1)   \quad K_1 = \frac{K}{mc}$

- $ \frac{d T_2}{T_2-T_1}=-\frac{k}{m c} d t=-K_1 d t$

- $\log _e(T_2-T_1)=-k_1 t+C $
 
 - $ T_2=T_1+C^{\prime} e^{-k_1 t}$
 
 - $C^{\prime}=e^c$

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<footer style = "font-size: 18px">Radiation $\rightarrow$ Stefan–Boltzmann Law of Radiation $\rightarrow$ <span style = "color:blue"> Newton's law of cooling </span> $\rightarrow$ Thank You $\rightarrow$  </footer>