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Thermal Properties of Matter MCQs for NEET — Physics Questions with Answers

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A liquid takes 5 minute to cool from $ 80 ^\circ C to 50 ^\circ C$ . The temperature of the surrounding is $ 20 ^\circ C$ . What is the time it will take to cool from $ 60 ^\circ C to 30 ^\circ C$ ?

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Explanation

Using the equation $ { \theta_1 - \theta _2 \over t } = K \left( {\theta_1 + \theta_2 \over 2} - \theta_0 \right) \theta_0$ = where = temperature of surrounding

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Two spheres of the same material have radii 1 m and 4 m and temperatures 2000 k and 4000 k respectively. If the energy radiated by the spheres are $E_1 $ and $E_2 $ respectively then find ratio of $ { E_1 \over E_2 } $

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Explanation

$ { E_1 \over E_2 } = \left( { R_1 \over R_2 } \right)^2 \left( { T_1 \over T_4 } \right)^4$

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A body cools in 5 minute from $ 60 ^\circ C to 40 ^\circ C $ .The temperature of the surroundings is $ 10 ^\circ C$ . What is its temperature after the next 5 minute ?

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Explanation

According to newton's law approximately, $ { \theta_1 - \theta _2 \over t } = K \left( {\theta_1 + \theta_2 \over 2} - \theta_0 \right) $ $ 160 - 4 \theta = 20 + \theta \Rightarrow \theta = 28 C $

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What is the units of emissive power in stefan's law ?

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Explanation

Emissive power is defined as the radiant energy emitted per sec per unit area of the surface. Hence $ [E} = [ P/A] \Rightarrow unit of E = Wm^{-2 } $

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A sphere, a cube and a thin circular plate are allmade of the same material, have the same mass and are initially heated to a temperature of 200°C.Arrange then in the ascending order of rate of cooling.

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Explanation

$ { d \theta \over dt } \alpha A $now, for a given mass,sphere has the least surface area and circular plate will have the maximum surface area. Hence the sphere will cool the slowest and the disc the fastest.

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100 g ice at 0°C placed in 100 g water at 100°C. The final temperature of the mixture will be.......... (Latent heat of ice is 80 Cal/g, and specific heat of water is 1 Cal/g C°)

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Explanation

Let temperature of mixture = T $ \therfore $ Heat absorbed by ice = Heat lost by water $ \therefore $ Heat required to melt ice + heat required to acquire Temperature T of water of ice = Heat lost by water $\therefore $ mL + mc (T - 0) = mc (100 - T)

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On a hot day at Ahmedabad a trucker loaded 37,000 L of diesel fuel. He delivered the disel at Shrinagar (Kashmir) Where the temperature was lower then that of Ahmedabad by 23 k. How many liters did he deliver ?For diesel $ \gamma = 3 \alpha = 9.50 \times 10^{-4} C° $ (Neglect the thermal expansion / Contration of steel tank of the trunk)

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Explanation

$ V = 37000 L \triangle T = 27K$. As temperature decreases in volume. $ \triangle V =3 \alpha V \triangle T= 9.5 \times 10^{-4} \times \times 37000 \times 23 $ = 808 L decreases $ \therefore $ Diesel supplied to shrinagar = 37000 - 808 = 36192L = 36190L

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For which value of the temperature will the values of Fahrenhit scale and Kelvin scale be equal ?

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Explanation

$T_F = {9 \over 15 } [T_K -273 ] + 32$

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The temperautre of equal masses of three different liquids x, y, z are $ 12 ^\circ C , 19 ^\circ C and 28 ^\circ C$ respectively. The temperature when X and Y are mixed is $ 16 ^\circ C$ and when Yand Z are mixed is $23 ^\circ C$ . what is the temperature when X and Z are mixed ?

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Explanation

When X and Y are mixed heat lost by Y = Heat gained by X $ms_y (19-16) = ms_X(16-12)$

Similarly when Y and Z are mixed $ ms_y (23-19) = ms_z (28-23)$ Mixing x and z yields, $ms_X(T -12) = ms_z (28 -T)$ Mixing X and Z yields. Solved all equations

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The Coefficient of linear expansion of brass & steel are $ \alpha_1 and \alpha_2 $ . 2 If we take a brass rod of length $ l_1 $ & steel rod of length $l_2 $ at $ 0 ^\circ $ , their difference in length $ ( l_2 - l_1 ) $ will remain the same at a temperature if ........................

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Explanation

$ l_2 = l_2 ( 1 + \alpha_2 \triangle Q) and L_1 = l_1 ( 1 + \alpha_1 \triangle Q ) $ $ \Rightarrow (l_2 - l_1) = ( l_2 -l_1 ) + \triangle Q ( l_2 \alpha_2 - l_1 \alpha _1 ) $ $ now (L_2 - L_1 ) = ( l_2 - l_1 ) $ $ so , l_2 \alpha_2 = l_2 \alpha_1 = 0 $

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