(A) Let the mass of the first part be $m$ and the mass of the second part be $(M - m)$.Heat released by $m \ kg$ of water at $t^{\circ} C$ to become i

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(A) Let the mass of the first part be $m$ and the mass of the second part be $(M - m)$.Heat released by $m \ kg$ of water at $t^{\circ} C$ to become ice at $0^{\circ} C$ is:$Q_1 = m \times c \times (t - 0) + m \times L_f = m \times 1 \times t + m \times 80 = m(t + 80)$.Heat absorbed by $(M - m) \ kg$ of water at $t^{\circ} C$ to become steam at $100^{\circ} C$ is:$Q_2 = (M - m) \times c \times (100 - t) + (M - m) \times L_v = (M - m) \times 1 \times (100 - t) + (M - m) \times 540 = (M - m)(640 - t)$.Equating $Q_1 = Q_2$:$m(t + 80) = (M - m)(640 - t)$$mt + 80m = 640M - Mt - 640m + mt$$80m + 640m = 640M - Mt$$720m = M(640 - t)$$\frac{m}{M} = \frac{640 - t}{720}$.

Class 11 Physics 10-1.Thermometry, Thermal Expansion and Calorimetry Mix Examples-Thermometry, Thermal Expansion and Calorimetry

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$M \ kg$ of water at $t^{\circ} C$ is divided into two parts so that one part of mass $m \ kg$ when converted into ice at $0^{\circ} C$ would release enough heat to vaporize the other part. Then $\frac{m}{M}$ is equal to (Specific heat of water $= 1 \ cal \ g^{-1} {}^{\circ} C^{-1}$,Latent heat of fusion of ice $= 80 \ cal \ g^{-1}$,Latent heat of steam $= 540 \ cal \ g^{-1}$)

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A
$\frac{640 - t}{720}$
B
$\frac{720 - t}{640}$
C
$\frac{640 + t}{720}$
D
$\frac{720 + t}{640}$

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10-1.Thermometry, Thermal Expansion and Calorimetry

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Mix Examples-Thermometry, Thermal Expansion and Calorimetry

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The heat required to change $1 \,kg$ of ice at $-8^{\circ} C$ into water at $20^{\circ} C$ at $1 \,atm$ of pressure is closest to .............$\,kJ$. (Assume that ice has a specific heat capacity $2.1 \,kJ / kg \cdot K$,water has a specific heat capacity $4.2 \,kJ / kg \cdot K$,and the latent heat of fusion of ice is $333 \,kJ / kg$.)

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$A$ cylinder of mass $1\,kg$ is given heat of $20000\,J$ at atmospheric pressure. If the initial temperature of the cylinder is $20\,^oC$,then the work done by the cylinder will be .......$J$. (Given: Specific heat of cylinder $= 400\,J\,kg^{-1}\,^oC^{-1}$,Coefficient of volume expansion $= 9 \times 10^{-5}\,^oC^{-1}$,Atmospheric pressure $= 10^5\,N/m^2$,and density of cylinder $= 9000\,kg/m^3$)

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$A$ metal ball immersed in alcohol weighs $W_1$ at $0^{\circ}C$ and $W_2$ at $59^{\circ}C$. The coefficient of cubical expansion of the metal is less than that of alcohol. Assuming that the density of metal is large compared to that of alcohol,it can be shown that

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$A$ lead ball moving with a velocity $V$ strikes a wall and stops. If $50\%$ of its energy is converted into heat,then what will be the increase in temperature? (Specific heat of lead is $S$)

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Column $I$ gives some devices and Column $II$ gives some processes on which the functioning of these devices depends. Match the devices in Column $I$ with the processes in Column $II$.
Column $I$ Column $II$
$(A)$ Bimetallic strip $(p)$ Radiation from a hot body
$(B)$ Steam engine $(q)$ Energy conversion
$(C)$ Incandescent lamp $(r)$ Melting
$(D)$ Electric fuse $(s)$ Thermal expansion of solids

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Column $I$	Column $II$
$(A)$ Bimetallic strip	$(p)$ Radiation from a hot body
$(B)$ Steam engine	$(q)$ Energy conversion
$(C)$ Incandescent lamp	$(r)$ Melting
$(D)$ Electric fuse	$(s)$ Thermal expansion of solids

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$A$ metal ball immersed in alcohol weighs $W_1$ at $0^{\circ}C$ and $W_2$ at $59^{\circ}C$. The coefficient of cubical expansion of the metal is less than that of alcohol. Assuming that the density of metal is large compared to that of alcohol,it can be shown that

$A$ lead ball moving with a velocity $V$ strikes a wall and stops. If $50\%$ of its energy is converted into heat,then what will be the increase in temperature? (Specific heat of lead is $S$)

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