$50\, g$ of ice at $0^{\circ}C$ is mixed with $50\, g$ of water at $80^{\circ}C$. The final temperature of the mixture will be........ $^{\circ}C$.

$A$ small quantity,mass $m$,of water at a temperature $\theta$ (in $^oC$) is poured onto a large mass $M$ of ice which is at its melting point. If $c$ is the specific heat of water and $L$ is the latent heat of fusion of ice,then the mass of ice melted is given by

Steam at $100^{\circ}C$ is passed into $1.1 \, kg$ of water contained in a calorimeter of water equivalent $0.02 \, kg$ at $15^{\circ}C$ until the temperature of the calorimeter and its contents rises to $80^{\circ}C$. The mass of the steam condensed in $kg$ is:

$A$ $2 \, kg$ block of ice at $-20^{\circ}C$ is added to $5 \, kg$ of water at $20^{\circ}C$. What will be the total mass of water in $kg$? (Specific heat of water = $1 \, kcal/kg/^{\circ}C$,specific heat of ice = $0.5 \, kcal/kg/^{\circ}C$,latent heat of fusion of ice = $80 \, kcal/kg$)

$A$ liquid of mass $m$ and specific heat $c$ is heated to a temperature $2T$. Another liquid of mass $m/2$ and specific heat $2c$ is heated to a temperature $T$. If these two liquids are mixed, the resulting temperature of the mixture is:

In an experiment,a sphere of aluminium of mass $0.20\, kg$ is heated up to $150\, ^\circ C$. Immediately,it is put into water of volume $150\, cc$ at $27\, ^\circ C$ kept in a calorimeter of water equivalent $0.025\, kg$. The final temperature of the system is $40\, ^\circ C$. The specific heat of aluminium is ............ $J/kg\cdot ^\circ C$ (take $4.2\, J = 1\, cal$).