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 $; J$)

For a copper-iron $(Cu-Fe)$ couple,the thermo $e.m.f.$ (temperature of cold junction $= 0\,^{\circ}C$) is given by $E = (14\theta - 0.02\theta^2)\,\mu V$. The neutral temperature will be

The height of a waterfall is $84 \, m$. Assuming that the entire kinetic energy of falling water is converted into heat, the rise in temperature of the water will be $(g = 9.8 \, m/s^2, J = 4.2 \, J/cal, c = 1 \, cal/g^\circ C = 4200 \, J/kg^\circ C)$. (in $^\circ C$)

The $e.m.f.$ in a thermoelectric circuit with one junction at $0\,^\circ C$ and the other at $t\,^\circ C$ is given by $E = At - Bt^2$. The neutral temperature is then

In cold countries,water pipes sometimes burst,because

An aluminium container of mass $100 \, g$ contains $200 \, g$ of ice at $-20^{\circ} C$. Heat is added to the system at the rate of $100 \, cal/s$. The temperature of the system after $4$ minutes will be ....... $^{\circ} C$ (specific heat of ice $= 0.5 \, cal/g^{\circ} C$,latent heat of fusion $L = 80 \, cal/g$,specific heat of $Al = 0.2 \, cal/g^{\circ} C$,specific heat of water $= 1 \, cal/g^{\circ} C$)