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, 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 of evaporation to specific heat of water is $540^{\circ} C$

  • [KVPY 2018]
  • A

    $2.7$

  • B

    $4.2$

  • C

    $5.4$

  • D

    $10.8$

Similar Questions

Heat is being supplied at a constant rate to the sphere of ice which is melting at the rate of $0.1 \,gm / s$. It melts completely in $100 \,s$. The rate of rise of temperature thereafter will be ............ $^{\circ} C / s$

The heat required to change $1 \,kg$ of ice at $-8^{\circ} C$ into water at $20^{\circ} C$. at $1 \,atm$ of pressure, is closest to .............$\,kJ$ (Assume that ice has a specific heat capacity $2.1 \,kJ / kg / K$. water has a specific heat capacity $4.2 \,kJ / kg / K$, and latent heat of fusion of ice is $333 \,kJ / kg$ )

  • [KVPY 2021]

A liquid at $30^{\circ} C$ is poured very slowly into a Calorimeter that is at temperature of $110^{\circ} C$. The boiling temperature of the liquid is $80^{\circ} C$. It is found that the first $5 gm$ of the liquid completely evaporates. After pouring another $80 gm$ of the liquid the equilibrium temperature is found to be $50^{\circ} C$. The ratio of the Latent heat of the liquid to its specific heat will be. . . . .${ }^{\circ} C$.  [Neglect the heat exchange with surrounding]

  • [IIT 2019]

The specific heat of water $=4200\, J\, kg ^{-1}\, K ^{-1}$ and the latent heat of ice $=3.4 \times 10^{5}\, J\, kg ^{-1}.$ $100$ grams of ice at $0^{\circ} C$ is placed in $200\, g$ of water at $25^{\circ} C$. The amount of ice that will melt as the temperature of water reaches $0^{\circ} C$ is close to (in $grams$) 

  • [JEE MAIN 2020]

Find the amount of heat supplied to decrease the volume of an ice water mixture by $1 \,\,cm^3$ without any change in temperature. $(\rho_ {ice} = 0.9 \rho_{water}, L_{ice} = 80 \,\,cal/gm).$ ......... $cal$