$0^o C$ $1\ gm$ ice is mixed with $100^o C$ vapor then what wiil be the final temperature ($^o C$)?
$100$
$230$
$130$
$50$
$50\, gm$ of copper is heated to increase its temperature by $10\,^oC$. If the same quantity of heat is given to $10\, gm$ of water, the rise in its temperature is ........ $^oC$ (Specific heat of copper $=420\, Joule\, kg^{-1}\,^oC^{-1}$)
The coefficient of apparent expansion of mercury in a glass vessel is $153 × 10^{-6}{°C^{-1}}$ and in a steel vessel is $144 × 10^{-6}{°C^{-1}}$. If $\alpha$ for steel is $12 × 10^{-6}{°C^{-1}}$, then that of glass is
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 $20\,^oC$. If $440\,g$ of hot water at $92\,^oC$ is poured in it, the final temperature (neglecting radiation loss) will be nearest to ........ $^oC$
The graph $AB$ shown in figure is a plot of temperature of a body in degree Celsius and degree Fahrenheit. Then
If the earth suddenly stops revolving and all its rotational $KE$ is used up in raising its temperature and if $'s'$ is taken to be the specific heat of the earth's material, the rise of temperature of the earth will be : ( $R -$ radius of the earth and $\omega =$ its angular velocity, $J =\,Joule$ constant)