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

Heat Quiz 2

physics

 A vessel full of specific liquid was heated up to a temperature $36{}^\circ C$. The initial temperature of the liquid was mentioned as $16{}^\circ C$ and the mass of the liquid inside the vessel was $400\text{ g}$. Calculate the measure of heat if the specific heat of capacitance is $0.32\text{ Jkg}{{\text{K}}^{-1}}$.

 The heat can be found by taking the product of mass, specific heat capacity and the change in temperature.

  The equation of heat be, $Q=mC\Delta T$ $ \\ $ Substituting the values in it, $ \\ Q=400\times {{10}^{-3}}\times 0.32\times \left( 36-16 \right) \\ \Rightarrow Q=0.4\times 0.32\times 20 \\ \therefore Q=2.56\text{ }J$

  The conductivity of the three states of matter i.e. solid, liquid and gas in descending order is,

 Liquids and gases are fluids that continuously flow and in solids the molecules are closely packed, the more the intermolecular spacing is less effective is the thermal energy transfer.

  Liquids and gases are fluids that continuously flow and in solids the molecules are closely packed, the more the intermolecular spacing is less effective is the thermal energy transfer. Solids have the least the intermolecular spacing and hence they have the greatest conductivity among the three states of matter, liquids have conductivity smaller than solids and in gases molecules have the most freedom i.e. the intermolecular spacing is largest in gas among the three states of matter, so they have the least conductivity.

  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 and to the distance between the hot and cold surface of the body. So, put in the values given to calculate the thermal conductivity.

 Conduction takes place when heat flows from a hot body to a cold body. Conductivity of a material is the degree to which it conducts heat. The conductive heat transfer is given by, $ \\ Q=\dfrac{K\cdot A\cdot ({{T}_{H}}-{{T}_{C}})}{l}    ….(i) \\ $ 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 \\ $  Putting the above values in equation $(i)$ and solving for $K$, 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=2.3\,J{{s}^{-1}}{{m}^{1}}{{K}^{-1}}$

 A conducting rod of length $2$m and cross section area $0.4\,{{m}^{2}}$and thermal conductivity $79.5\,W{{m}^{-1}}{{K}^{-1}}$, has the temperature difference between it two ends as $87\,K$, then what is the heat transfer per unit time in this rod ?

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

 Conductive heat transfer is given by, $ \\ Q=\dfrac{K\cdot A\cdot \Delta T}{l}   ….(i) \\ $ According to the question, $ \\ K=79.5W{{m}^{-1}}{{K}^{-1}},A=0.4{{m}^{2}},l=2m,\Delta T={{87}^{0}}K \\ $ Putting the above values in equation $(i)$, we get, $ \\ Q=\dfrac{K\cdot A\cdot \Delta T}{l} \\ \therefore Q=\dfrac{79.5\times 0.4\times 87}{2}=1.4\,kW$

 What will be the power radiated from a black body of surface area $3.5\,{{m}^{2}}$ at temperature $396\,K$ of emissivity $0.46$ and the surrounding temperature is $235\,K$ ? $($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.

 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) \\ $ 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 \\ $ 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$