$A$ waterfall is $84 \ m$ high. If half of the potential energy of the falling water gets converted to heat,the rise in temperature of the water will be: (in $^\circ C$)

$A$ thermoelectric emf of $200 \, \mu V$ is generated between $0 \, ^oC$ and $100 \, ^oC$. The emf generated between $(0 \, ^oC - 32 \, ^oC)$ and $(32 \, ^oC - 70 \, ^oC)$ are $64 \, \mu V$ and $76 \, \mu V$ respectively. What is the thermo $emf$ generated between $(70 \, ^oC - 100 \, ^oC)$ in $\mu V$?

Find the quantity of heat required to convert $40 \; g$ of ice at $-20^{\circ} C$ into water at $20^{\circ} C$. Given $L_{\text{ice}} = 0.336 \times 10^6 \; J/kg$,specific heat of ice $= 2100 \; J/kg \cdot K$,and specific heat of water $= 4200 \; J/kg \cdot K$. (in $; J$)

$M \ kg$ of water at $t^{\circ} C$ is divided into two parts so that one part of mass $m \ kg$ when converted into ice at $0^{\circ} C$ would release enough heat to vaporize the other part. Then $\frac{m}{M}$ is equal to (Specific heat of water $= 1 \ cal \ g^{-1} {}^{\circ} C^{-1}$,Latent heat of fusion of ice $= 80 \ cal \ g^{-1}$,Latent heat of steam $= 540 \ cal \ g^{-1}$)

$A$ 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 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 $333 \,kJ / kg$.)