A piece of ice (heat capacity $=2100 \mathrm{~J} \mathrm{~kg}^{-1}{ }^{\circ} \mathrm{C}^{-1}$ and latent heat $=3.36 \times 10^5 \mathrm{~J} \mathrm{~kg}^{-1}$ ) of mass $\mathrm{m}$ grams is at $-5^{\circ} \mathrm{C}$ at atmospheric pressure. It is given $420 \mathrm{~J}$ of heat so that the ice starts melting. Finally when the ice-water mixture is in equilibrium, it is found that $1 \ \mathrm{gm}$ of ice has melted. Assuming there is no other heat exchange in the process, the value of $m$ is
$7$
$8$
$9$
$5$
$50\, g$ ice at $0\,^oC$ is dropped into a calorimeter containing $100\, g$ water at $30\,^oC$. If thermal capacity of calorimeter is zero then amount of ice left in the mixture at equilibrium is ........ $gm$
A kettle with $2\, littre$ water at $27\,^oC$ is heated by operating coil heater of power $1\, kW$. The heat is lost to the atmosphere at constant rate $160\, J/sec$, when its lid is open. In how much time will water heated to $77\,^oC$. (specific heat of water $= 4.2\, kJ/kg$) with the lid open ?
A water cooler of storage capacity $120$ litres can cool water at a constant rate of $P$ watts. In a closed circulation system (as shown schematically in the figure), tr e wat'r from the cooler is used to cool an external device that generates constantly $3 \mathrm{~kW}$ of heat (thermal load). The temperature of water fed into the device cannot exceed $30^{\circ} \mathrm{C}$ and the e.tire stored $120$ litres of water is initially cooled to $10^{\circ} \mathrm{C}$. The entire system is thermally insulat $\mathrm{d}$. The minimum value of $P$ (in watts) for which the device can be operated for $3$ hours is
(Specific heat of water is $4.2 \mathrm{~kJ}^{-1} \mathrm{~kg}^{-1}$ and the density of water is $10.$) $0 \mathrm{k}^2 \mathrm{~m}^{-3}$ )
The thermal capacity of a body is $80\, cal$, then its water equivalent is
An electric kettle (rated accurately at $2.5\, kW$) is used to heat $3\, kg$ of water from $15\,^oC$ to boiling point. It takes $9.5$ minute. Then, the amount of heat that has been lost is