The thermo $e.m.f.$ of a thermocouple varies with the temperature $\theta$ of the hot junction as $E = a\theta + b\theta^2$ in volts,where the ratio $a/b$ is $700\,^{\circ}C$. If the cold junction is kept at $0\,^{\circ}C$,then the neutral temperature is:

$A$ child running a temperature of $101\,^{\circ} F$ is given an antipyrin (i.e. a medicine that lowers fever) which causes an increase in the rate of evaporation of sweat from his body. If the fever is brought down to $98\,^{\circ} F$ in $20$ minutes,what is the average rate of extra evaporation caused by the drug (in $g/min$)? Assume the evaporation mechanism to be the only way by which heat is lost. The mass of the child is $30\; kg$. The specific heat of the human body is approximately the same as that of water,and the latent heat of evaporation of water at that temperature is about $580\; cal \;g^{-1}$.

$A$ metal block is made from a mixture of $2.4 \ kg$ of aluminium,$1.6 \ kg$ of brass,and $0.8 \ kg$ of copper. The metal block is initially at $20^{\circ} C$. If the heat supplied to the metal block is $44.4 \ cal$,find the final temperature of the block if the specific heats of aluminium,brass,and copper are $0.216, 0.0917, 0.0931 \ cal \cdot kg^{-1} \cdot ^{\circ}C^{-1}$ respectively. (in $^{\circ} C$)

On which of the following four processes does the functioning of a bimetallic strip depend?
$(i)$ Radiation
$(ii)$ Energy conversion
$(iii)$ Melting
$(iv)$ Thermal expansion

An earthen pitcher used in summer cools water in it essentially by evaporation of water from its porous surface. If a pitcher carries $4 \,kg$ of water and the rate of evaporation is $20 \,g$ per hour,the temperature of water in it decreases by $\Delta T$ in two hours. The value of $\Delta T$ is close to ........... $^{\circ} C$ (ratio of latent heat of evaporation to specific heat of water is $540^{\circ} C$).

$A$ gun fires a lead bullet of temperature $300 \ K$ into a wooden block. The bullet having a melting temperature of $600 \ K$ penetrates into the block and melts down. If the total heat required for the process is $625 \ J$,then the mass of the bullet is . . . . . . grams. (Latent heat of fusion of lead $= 2.5 \times 10^4 \ J \ kg^{-1}$ and specific heat capacity of lead $= 125 \ J \ kg^{-1} K^{-1}$)