$A$ $20 \,g$ bullet whose specific heat is $5000 \,J/(kg \cdot ^{\circ}C)$ and moving at $2000 \,m/s$ plunges into a $1.0 \,kg$ block of wax whose specific heat is $3000 \,J/(kg \cdot ^{\circ}C)$. Both the bullet and the wax are at $25^{\circ}C$. Assuming that $(i)$ the bullet comes to rest in the wax and $(ii)$ all its kinetic energy is converted into heat,the final temperature of the wax (in $^{\circ}C$) is close to:

$A$ calorimeter has a water equivalent of $5 \times 10^{-3} \ kg$ and contains $25 \times 10^{-3} \ kg$ of water. It takes $3 \ minutes$ to cool from $28^{\circ}C$ to $21^{\circ}C$. When it is filled with turpentine oil,it takes $2 \ minutes$ to cool from $28^{\circ}C$ to $21^{\circ}C$. The specific heat of turpentine oil is ..... $cal/g^{\circ}C$.

The heat required to convert $1 \, g$ of ice at $0^{\circ}C$ into steam at $100^{\circ}C$ is (given $L_{f} = 80 \, cal/g$,$c_{w} = 1 \, cal/g^{\circ}C$,$L_{v} = 536 \, cal/g$):

An amount of ice of mass $10^{-3} \ kg$ and temperature $-10^{\circ} C$ is transformed to vapour of temperature $110^{\circ} C$ by applying heat. The total amount of heat required for this conversion is,(Take,specific heat of ice $= 2100 \ J \ kg^{-1} \ K^{-1}$,specific heat of water $= 4180 \ J \ kg^{-1} \ K^{-1}$,specific heat of steam $= 1920 \ J \ kg^{-1} \ K^{-1}$,Latent heat of ice $= 3.35 \times 10^5 \ J \ kg^{-1}$ and Latent heat of steam $= 2.25 \times 10^6 \ J \ kg^{-1}$) (in $J$)

The heat required to convert $8 \ g$ of ice at a temperature of $-20^{\circ}C$ to steam at $100^{\circ}C$ is (Specific heat capacity of ice $= 2100 \ J \ kg^{-1} \ K^{-1}$, specific heat capacity of water $= 4200 \ J \ kg^{-1} \ K^{-1}$, latent heat of fusion of ice $= 336 \times 10^3 \ J \ kg^{-1}$ and latent heat of steam $= 2.268 \times 10^6 \ J \ kg^{-1}$) (in $kJ$)

$A$ bullet of mass $5\, g$,travelling with a speed of $210\, m/s$,strikes a fixed wooden target. One half of its kinetic energy is converted into heat in the bullet while the other half is converted into heat in the wood. The rise of temperature of the bullet if the specific heat of its material is $0.030\, cal/(g \cdot ^{\circ}C)$ $(1\, cal = 4.2 \times 10^{7}\, ergs)$ is close to $.......^{\circ}C$.