(A) The volume of mercury spilled is given by the difference in the expansion of mercury and the expansion of the glass flask.The formula for the volu

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(A) The volume of mercury spilled is given by the difference in the expansion of mercury and the expansion of the glass flask.The formula for the volume of liquid spilled is $\Delta V = V_0 (\gamma_m - \gamma_g) \Delta T$,where $\gamma_g = 3 \alpha_g$.Given:$V_0 = 1 \text{ litre} = 1000 \text{ ml}$$\gamma_m = 1.82 \times 10^{-4} /^{\circ}C$$\alpha_g = 10 \times 10^{-6} /^{\circ}C \implies \gamma_g = 3 \times 10 \times 10^{-6} = 0.3 \times 10^{-4} /^{\circ}C$$\Delta T = 100 - 0 = 100^{\circ}C$Substituting the values:$\Delta V = 1000 \times (1.82 \times 10^{-4} - 0.3 \times 10^{-4}) \times 100$$\Delta V = 1000 \times (1.52 \times 10^{-4}) \times 100$$\Delta V = 1000 \times 0.0152 = 15.2 \text{ ml}$.

Class 11 Physics 1 Thermal Properties of Matter: Thermometry, Expansion and Calorimetry Mix Examples-Thermometry, Thermal Expansion and Calorimetry

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$A$ glass flask of volume $1$ litre is fully filled with mercury at $0\,^{\circ}C$. Both the flask and mercury are now heated to $100\,^{\circ}C$. If the coefficient of volume expansion of mercury is $1.82 \times 10^{-4}/^{\circ}C$ and the coefficient of linear expansion of glass is $10 \times 10^{-6}/^{\circ}C$,the amount of mercury which is spilled out is ............ $ml$. (in $.2$)

A
$15$
B
$17$
C
$19$
D
$21$

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1 Thermal Properties of Matter: Thermometry, Expansion and Calorimetry

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Mix Examples-Thermometry, Thermal Expansion and Calorimetry

The height of a waterfall is $84 \, m$. Assuming that the entire kinetic energy of falling water is converted into heat, the rise in temperature of the water will be $(g = 9.8 \, m/s^2, J = 4.2 \, J/cal, c = 1 \, cal/g^\circ C = 4200 \, J/kg^\circ C)$. (in $^\circ C$)

Medium

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One $kg$ of water, at $20, ^oC$, is heated in an electric kettle whose heating element has a mean (temperature averaged) resistance of $20, \Omega$. The rms voltage in the mains is $200, V$. Ignoring heat loss from the kettle, the time taken for water to evaporate fully is close to.......... $min$. [Specific heat of water $= 4200, J/kg, ^oC$, Latent heat of water $= 2260, kJ/kg$]

DifficultJEE MAIN 2019

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$A$ block of ice at $0^{\circ}C$ falls from a height of $1 \ km$ onto an insulating surface. If all its kinetic energy is converted into heat,what fraction of the ice will melt? (Take $g = 10 \ m/s^{2}$ and latent heat of fusion of ice $L = 3.36 \times 10^{5} \ J/kg$)

Medium

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If liquefied oxygen at $1$ atmospheric pressure is heated from $50\, K$ to $300\, K$ by supplying heat at a constant rate,the graph of temperature vs. time will be:

MediumAIIMS 2012

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$A$ bullet of mass $m$ moving with a uniform velocity $v$ stops suddenly after hitting a target. If the whole mass melts,given the specific heat $S$,initial temperature $25^{\circ}C$,melting point $475^{\circ}C$,and latent heat $L$,then $v$ is given by:

Medium

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The height of a waterfall is $84 \, m$. Assuming that the entire kinetic energy of falling water is converted into heat, the rise in temperature of the water will be $(g = 9.8 \, m/s^2, J = 4.2 \, J/cal, c = 1 \, cal/g^\circ C = 4200 \, J/kg^\circ C)$. (in $^\circ C$)

$A$ block of ice at $0^{\circ}C$ falls from a height of $1 \ km$ onto an insulating surface. If all its kinetic energy is converted into heat,what fraction of the ice will melt? (Take $g = 10 \ m/s^{2}$ and latent heat of fusion of ice $L = 3.36 \times 10^{5} \ J/kg$)

If liquefied oxygen at $1$ atmospheric pressure is heated from $50\, K$ to $300\, K$ by supplying heat at a constant rate,the graph of temperature vs. time will be:

$A$ bullet of mass $m$ moving with a uniform velocity $v$ stops suddenly after hitting a target. If the whole mass melts,given the specific heat $S$,initial temperature $25^{\circ}C$,melting point $475^{\circ}C$,and latent heat $L$,then $v$ is given by:

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