When $M_1$ gram of ice at $-10\,^{\circ}C$ (specific heat $= 0.5\, cal\, g^{-1}\,^{\circ}C^{-1}$) is added to $M_2$ gram of water at $50\,^{\circ}C$,finally no ice is left and the water is at $0\,^{\circ}C$. The value of latent heat of ice,in $cal\, g^{-1}$ is

  • A
    $\frac{50M_2}{M_1} - 5$
  • B
    $\frac{5M_2}{M_1} - 5$
  • C
    $\frac{50M_2}{M_1}$
  • D
    $\frac{5M_1}{M_2} - 50$

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$A$ beaker contains $200\,g$ of water. The heat capacity of the beaker is equal to that of $20\,g$ of water. The initial temperature of water in the beaker is $20\,^{\circ}C$. If $440\,g$ of hot water at $92\,^{\circ}C$ is poured in it,the final temperature (neglecting radiation loss) will be nearest to ........ $^{\circ}C$.

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$100 \text{ g}$ of ice at $0^{\circ}C$ is mixed with $100 \text{ g}$ of water at $100^{\circ}C$. The final temperature of the mixture is. [Take,$L_f = 3.36 \times 10^5 \text{ J kg}^{-1}$ and $S_w = 4.2 \times 10^3 \text{ J kg}^{-1} \text{ K}^{-1}$] (in $^{\circ}C$)

An aluminium piece of mass $50 \,g$ initially at $300^{\circ} C$ is dipped quickly and taken out of $1 \,kg$ of water,initially at $30^{\circ} C$. If the temperature of the aluminium piece immediately after being taken out of the water is found to be $160^{\circ} C$,the temperature of the water is ............ $^{\circ} C$. The specific heat capacities of aluminium and water are $900 \,J \,kg^{-1} K^{-1}$ and $4200 \,J \,kg^{-1} K^{-1}$,respectively.

$A$ liquid of specific heat $0.8 \ cal / g^{\circ} C$ at temperature $60^{\circ} C$ is mixed with another liquid of the same mass having temperature $45^{\circ} C$. If the temperature of the mixture is $53^{\circ} C$,then the specific heat (in $cal / g^{\circ} C$) of the second liquid is:

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