Steam of mass $60 \ g$ at a temperature $100^{\circ} C$ is mixed with water of mass $360 \ g$ at a temperature $40^{\circ} C$. The ratio of the masses of steam and water in equilibrium is (Latent heat of steam is $540 \ cal \ g^{-1}$ and specific heat capacity of water is $1 \ cal \ g^{-1} {}^{\circ} C^{-1}$)

$A$ coffee maker makes coffee by passing steam through a mixture of coffee powder,milk,and water. If the steam is mixed at the rate of $50 \, g/min$ in a mug containing $500 \, g$ of mixture,then it takes about $t_0$ seconds to make coffee at $70^{\circ} C$ when the initial temperature of the mixture is $25^{\circ} C$. The value of $t_0$ is close to .......... $s$ (ratio of latent heat of evaporation to specific heat of water is $540^{\circ} C$ and specific heat of the mixture can be taken to be the same as that of water).

$A$ $210 \,W$ 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 $\,s$)

Steam at $100^{\circ}C$ is added slowly to $1400 \, \text{g}$ of water at $16^{\circ}C$ until the temperature of the water is raised to $80^{\circ}C$. The mass of steam required to do this is $(L_V = 540 \, \text{cal/g})$ ........... $\text{g}$

$A$ piece of ice (heat capacity $= 2100 \ J \ kg^{-1} \ ^{\circ}C^{-1}$ and latent heat $= 3.36 \times 10^5 \ J \ kg^{-1}$) of mass $m$ grams is at $-5^{\circ}C$ at atmospheric pressure. It is given $420 \ J$ of heat so that the ice starts melting. Finally,when the ice-water mixture is in equilibrium,it is found that $1 \ g$ of ice has melted. Assuming there is no other heat exchange in the process,the value of $m$ is:

When $50 \ g$ of water at $10^{\circ} C$ is mixed with $50 \ g$ of water at $100^{\circ} C$,the resultant temperature is: (in $^{\circ} C$)