$A$ $210 \,W$ heater is used to heat $100 \,g$ of water. The time required to raise the temperature of this water from $25^{\circ} C$ to $100^{\circ} C$ is (specific heat capacity of water $= 4200 \,J / kg \cdot ^{\circ} C$). (in $\,s$)

$A$ calorimeter contains $0.2\, kg$ of water at $30\,^{\circ}C$. $0.1\, kg$ of water at $60\,^{\circ}C$ is added to it. The mixture is well stirred and the resulting temperature is found to be $35\,^{\circ}C$. The thermal capacity of the calorimeter is .......... $J/K$.

$A$ metal ball of mass $100 \ g$ at $20^{\circ} C$ is dropped in $200 \ ml$ of water at $80^{\circ} C$. If the resultant temperature is $70^{\circ} C$,then the ratio of specific heat of the metal to that of water is

When $M_1$ gram of ice at $-10\,^{\circ}C$ (specific heat $= 0.5\, cal\, g^{-1}\,^{\circ}C^{-1}$) is added to $M_2$ gram of water at $50\,^{\circ}C$,finally no ice is left and the water is at $0\,^{\circ}C$. The value of latent heat of ice,in $cal\, g^{-1}$ is

An iron ball of mass $0.2\,kg$ is heated to $100\,^{\circ}C$ and put into a block of ice at $0\,^{\circ}C.$ If $25\,g$ of ice melts,and the latent heat of fusion of ice is $80\,cal/g,$ find the specific heat of iron in $cal/g\,^{\circ}C.$

The water equivalent of a calorimeter is $10 \ g$ and it contains $50 \ g$ of water at $15^{\circ} C$. Some amount of ice,initially at $-10^{\circ} C$,is dropped in it and half of the ice melts till equilibrium is reached. What was the initial amount of ice that was dropped (given specific heat of ice $= 0.5 \ cal \ g^{-1} {}^{\circ} C^{-1}$,specific heat of water $= 1.0 \ cal \ g^{-1} {}^{\circ} C^{-1}$ and latent heat of melting of ice $= 80 \ cal \ g^{-1}$) (in $g$)?