$50 \, g$ of ice at $0^\circ C$ is placed in an insulated vessel. $50 \, g$ of water at $100^\circ C$ is mixed into it. Neglecting heat loss,what is the final temperature of the mixture?

$37 \ g$ of ice at $0^{\circ} C$ temperature is mixed with $74 \ g$ of water at $70^{\circ} C$ temperature. The resultant temperature is (Specific heat capacity of water $= 1 \ cal \ g^{-1} {}^{\circ} C^{-1}$ and latent heat of fusion of ice $= 80 \ cal \ g^{-1}$) (in $^{\circ} C$)

$A$ $2\,kg$ copper block is heated to $500^\circ C$ and then it is placed on a large block of ice at $0^\circ C$. If the specific heat capacity of copper is $400\, J/kg/ ^\circ C$ and latent heat of fusion of water is $3.5 \times 10^5\, J/kg$,the amount of ice that can melt is:

$A$ calorimeter contains $0.5 \,kg$ of water at $30^{\circ} C$. When $0.3 \,kg$ of water at $60^{\circ} C$ is added to it, the resulting temperature is found to be $40^{\circ} C$. The water equivalent of the calorimeter is (in $\,kg$)

Pure water supercooled to $-15^{\circ} C$ is contained in a thermally insulated flask. $A$ small amount of ice is thrown into the flask. The fraction of water frozen into ice is:

In an experiment on the specific heat of a metal,a $0.20 \; kg$ block of the metal at $150 \; ^{\circ}C$ is dropped in a copper calorimeter (of water equivalent $0.025 \; kg$) containing $150 \; cm^{3}$ of water at $27 \; ^{\circ}C$. The final temperature is $40 \; ^{\circ}C$. Compute the specific heat of the metal. If heat losses to the surroundings are not negligible,is your answer greater or smaller than the actual value for the specific heat of the metal?