$2 \; g$ of steam condenses when passed through $40 \; g$ of water initially at $25 \; ^\circ C$. The condensation of steam raises the temperature of water to $54.3 \; ^\circ C$. What is the latent heat of steam in $cal/g$?

$50 \, g$ of ice at $0^\circ C$ is placed in an insulated vessel. $50 \, g$ of water at $100^\circ C$ is mixed into it. Neglecting heat loss,what is the final temperature of the mixture?

$A$ calorimeter of water equivalent $20 \, g$ contains $180 \, g$ of water at $25^{\circ} C$. '$m$' grams of steam at $100^{\circ} C$ is mixed in it until the temperature of the mixture is $31^{\circ} C$. The value of '$m$' is close to (Latent heat of water $= 540 \, \text{cal} \, g^{-1}$,specific heat of water $= 1 \, \text{cal} \, g^{-1} {}^{\circ} C^{-1}$)

Three liquids with masses $m_1, m_2, m_3$ are thoroughly mixed. If their specific heats are $c_1, c_2, c_3$ and their temperatures are $T_1, T_2, T_3$ respectively,then the temperature of the mixture is

Objects $A$ and $B$ that are initially separated from each other and well isolated from their surroundings are then brought into thermal contact. Initially $T_A = 0^{\circ}C$ and $T_B = 100^{\circ}C$. The specific heat of $A$ is less than the specific heat of $B$. After some time,the system comes to an equilibrium state. The final temperatures are :

$A$ 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).