One quality of a thermometer is that its heat capacity should be small. If $P$ is a mercury thermometer,$Q$ is a resistance thermometer,and $R$ is a thermocouple type,then:

$A$ thermally insulated vessel contains $150\, g$ of water at $0\, ^oC$. Then the air from the vessel is pumped out adiabatically. $A$ fraction of water turns into ice and the rest evaporates at $0\, ^oC$ itself. The mass of evaporated water will be closest to ....... $g$ (Latent heat of vaporization of water $= 2.10 \times 10^6\, J/kg$ and Latent heat of fusion of water $= 3.36 \times 10^5\, J/kg$).

The heat required to change $1 \,kg$ of ice at $-8^{\circ} C$ into water at $20^{\circ} C$ at $1 \,atm$ of pressure is closest to .............$\,kJ$. (Assume that ice has a specific heat capacity $2.1 \,kJ / kg \cdot K$,water has a specific heat capacity $4.2 \,kJ / kg \cdot K$,and the latent heat of fusion of ice is $333 \,kJ / kg$.)

Water falls from a height of $200 \ m$ into a pool. Calculate the rise in temperature of the water assuming no heat dissipation from the water in the pool. ($g = 10 \ m/s^2$,specific heat of water $s = 4200 \ J/(kg \ K)$) (in $K$)

Consider a rectangular sheet of solid material of length $l=9 \ cm$ and width $d=4 \ cm$. The coefficient of linear expansion is $\alpha=3.1 \times 10^{-5} \ K^{-1}$ at room temperature and one atmospheric pressure. The mass of the sheet is $m=0.1 \ kg$ and the specific heat capacity is $C_v=900 \ J \ kg^{-1} K^{-1}$. If the amount of heat supplied to the material is $8.1 \times 10^2 \ J$,then the change in area of the rectangular sheet is:

$A$ lead bullet of $20 \ g$ travelling at $300 \ m/s$ strikes against a block of wood and comes to rest. Assuming $50 \%$ of heat is absorbed by the bullet,the increase in its temperature is $.....^{\circ} C$ (take specific heat of lead $= 150 \ J/kg \cdot K$).