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This is one of our many quizzes on Heat transfer. Click on this to access more such competitive quizzes on Thermal Physics.

Heat transfer Quiz 2

physics

Heat transfer $Q$, when convection is taking place is given by, 

 The heat transfer $Q$, when convection is taking place is directly proportional to $A$, the area and $\Delta T$, the temperature difference. The coefficient of proportionality is the convective heat transfer coefficient. 

 Convection takes place in liquids. The convective heat transfer per unit time of a liquid is directly proportional to $A$, the area and $\Delta T$, the temperature difference. The coefficient of proportionality is the convective heat transfer coefficient. Therefore the heat transfer $Q$, when convection is taking place is $Q={{h}_{c}}\cdot A\cdot ({{T}_{2}}-{{T}_{1}})$.Here, $h$ is the convective heat transfer coefficient, $A$ is the area and $\Delta T$ is the temperature difference.

An unknown metal of cross section area $2.2\,{{m}^{2}}$ and length $3\,m$ is heated. The hot end is at ${{58}^{0}}C$ and it's cold end is at ${{25}^{0}}C$ and the heat transfer per unit time is $56\,J{{s}^{-1}}$. What is the thermal conductivity of this metal? 

 The thermal conductivity, $K$ is directly proportional to $Q$, the conductive heat transfer and $l$, the distance between the hot and cold surface of the body but is inversely proportional to $A$, the area and $\Delta T$, the temperature difference.So, put in the values given to calculate the thermal conductivity. 

 $\text{Conduction takes place when heat flows from a hot body to a cold body.}\\ \text{Conductivity of a material is the degree to which it conducts heat.}\\ \text{The conductive heat transfer is given by,}\\ Q=\dfrac{K\cdot A\cdot ({{T}_{H}}-{{T}_{C}})}{l}….(i)\\ \text{According to the question,}\\ Q=56J{{s}^{-1}},A=2.2{{m}^{2}},l=3m,{{T}_{C}}={{25}^{0}}C,{{T}_{H}}={{58}^{0}}C\\ \text{Putting the above values in equation (i) and solving for K,}\\ \text{the thermal conductivity, we get,}\\ Q=\dfrac{K\cdot A\cdot ({{T}_{H}}-{{T}_{C}})}{l}\\ \Rightarrow K=\dfrac{Q\cdot l}{A\cdot ({{T}_{H}}-{{T}_{C}})}\\ \Rightarrow K=\dfrac{56\times 3}{2.2\times (58-25)}\\ \therefore K =2.3J{{s}^{-1}}{{m}^{1}}{{K}^{-1}}$

What will be the power radiated from a black body of surface area $3.5\,{{m}^{2}}$ at temperature $396K$ of emissivity $0.46$ and the surrounding temperature is $235K$. $(\text{Stefan's constant is}\,5.67\times {{10}^{-8}}\,W{{m}^{-2}}{{K}^{-4}})$ 

 The power radiated $P$, from a black body and its temperature is related as $P\propto {{T}^{4}}$, $T$ is the temperature. $P$ is given by $P=e\cdot \sigma \cdot A\cdot ({{T}^{4}}-T_{C}^{4})$. Put the values of emissivity, $e$ and area, $A$ and the temperature difference, given in the question in the above equation we can find the net power radiated from a black body. 

$\text{The net power radiated from a black body is given by the equation,}\\ P=e\cdot \sigma \cdot A\cdot ({{T}^{4}}-T_{C}^{4})….(i)\\ \text{According to the question,}\\ e=0.46,\sigma =5.67\times {{10}^{-8}}W{{m}^{-2}}{{K}^{-4}},A=3.5{{m}^{2}},T=396K,{{T}_{C}}=235K\\ \text{Putting the above values in equation (i), we get,}\\ P=0.46\times 5.67\times {{10}^{-8}}\times 3.5\times [{{(396)}^{4}}-{{(235)}^{4}}]\\ \therefore P =1.97\,kW$

 A rod made by attaching two metal $P$ and $Q$ of length $1$m and $1.5$m, respectively and cross section area of $P$ is twice of that of $Q$ and thermal conductivity of $P$ is $50.2\,J{{s}^{-1}}{{m}^{-1}}{{K}^{-1}}$ and of $Q$ is $385\,J{{s}^{-1}}{{m}^{-1}}{{K}^{-1}}$ end to end as shown in the figure. If this rod is kept between a hot body at ${{343}^{0}}K$ and cold body at ${{110}^{0}}K$, what is the temperature $T$ at the junction of $P$ and $Q$?<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/6915c231-8952-40e8-ba19-711588017bb8966683379649531396.png"/> 

 In the steady state, consider any cross-section of the rod, heat flowing into the element has to be equal to the heat flowing out of it. Hence, the rate of heat flow across a cross-section of the rod is the same at every point throughout the length of the combined rod. Equate the heat flow at the junction to find $T$. 

$\text{Conduction takes place when heat flows from a hot body to a cold body.}\\ \text{Conductivity of a material is the degree to which it conducts heat.}\\ \text{The conductive heat transfer is given by,}\\Q=\dfrac{K\cdot A\cdot ({{T}_{H}}-{{T}_{C}})}{l}….(i)\\ \text{Assuming there is no net loss in heat transfer from hot body to cold body in the steady state,}\\ \text{if we consider any cross-section of the rod, heat flowing into the element has to be equal to the heat flowing out of it.}\\ \text{Hence, the rate of heat flow across a cross-section of the rod is the same at every point}\\ \text{throughout the length of the combined rod. So,}\,{{Q}_{P}}={{Q}_{Q}}\,\text{at the junction,}\\ \dfrac{{{K}_{P}}\cdot {{A}_{P}}\cdot ({{T}_{H}}-T)}{{{l}_{P}}}=\dfrac{{{K}_{Q}}\cdot {{A}_{Q}}\cdot (T-{{T}_{C}})}{{{l}_{Q}}}….(ii)\\ \text{According to the question, T is the temperature of the junction, and,}\\ {{K}_{P}}=50.2J{{s}^{-1}}{{m}^{-1}}{{K}^{-1}},{{K}_{Q}}=385J{{s}^{-1}}{{m}^{-1}}{{K}^{-1}},{{A}_{P}}=2{{A}_{Q}}, \\ {{l}_{P}}=1m,{{l}_{Q}}=1.5m,{{T}_{H}}={{343}^{0}}K,{{T}_{C}}={{110}^{0}}K \\ \text{Putting the above values in equation (ii), we get,}\\ \dfrac{50.2\times {{A}_{P}}\times (343-T)}{1}=\dfrac{385\times 2{{A}_{P}}\times (T-110)}{1.5} \\ \therefore T={{97.2}^{0}}\,K$

 What is the emissivity of a black body radiating power equal to $2.3\,kW$ having surface area $9.2\,{{m}^{2}}$ at a temperature ${{287}^{0}}\,K$ and the surroundings at a temperature ${{80}^{0}}\,K$?$(\text{Stefan's constant is}\,5.67\times {{10}^{-8}}\,W{{m}^{-2}}{{K}^{-4}})$ 

 The power radiated $P$, from a black body is directly proportional to emissivity, $e$ as it is seen from the equation $P=e\cdot \sigma \cdot A\cdot ({{T}^{4}}-T_{C}^{4})$. To calculate the emissivity, $e$ we simply cross multiply the power radiated P and area, A and the temperature difference given in the question and solve.

$\text{The net power radiated from a black body is given by the equation,}\\ P=e\cdot \sigma \cdot A\cdot ({{T}^{4}}-T_{C}^{4})….(i)\\ \text{According to the question,}\\ P=2.3kW,\sigma =5.67\times {{10}^{-8}}W{{m}^{-2}}{{K}^{-4}},A=9.2{{m}^{2}},T={{287}^{0}}K,{{T}_{C}}={{80}^{0}}K\\ \text{Putting the above values in equation (i) and solving for e, we get,}\\ e=\dfrac{P}{\sigma \cdot A\cdot ({{T}^{4}}-T_{C}^{4})}\\ \Rightarrow e =\dfrac{2.3\times {{10}^{3}}}{5.67\times {{10}^{-8}}\times 9.2\times [{{(287)}^{4}}-{{(80)}^{4}}]}\\ \therefore e=0.7$