148215 A $10 \mathrm{H}$ inductor carries of $20 \mathrm{~A}$. If the energy stored in this inductor melts ice at $0^{\circ} \mathrm{C}$, then the mass of the ice melted is [Latent heat of ice $\left.=\mathbf{2 . 2 6} \times 10^{3} \mathrm{~J} \mathrm{~kg}^{-1}\right]$

148216 A small electric heater is used to heat $200 \mathrm{~g}$ of water. The time required to bring all this water from $40^{\circ} \mathrm{C}$ to $100^{\circ} \mathrm{C}$ is $200 \mathrm{~s}$. If specific heat of the water is $42 \mathrm{Jkg}^{-1} \mathrm{~K}^{-1}$ then the power supplied by the heater is-

148217 $5 \mathrm{~g}$ of ice at $-30{ }^{\circ} \mathrm{C}$ and $20 \mathrm{~g}$ of water at $35^{\circ} \mathrm{C}$ are mixed together in a calorimeter. The final temperature of the mixture is
(Neglect heat capacity of the calorimeter specific hear capacity of ice $=0.5 \mathrm{cal} \mathrm{g}^{-1}{ }^{0} \mathrm{C}^{-1}$ and latent heat of fusion of ice $=80 \mathrm{cal} \mathrm{g}^{-1}$ and specific hear capacity of water $=1 \mathrm{cal} \mathrm{g}^{-1}{ }^{\circ} \mathrm{C}^{-1}$ )

148218 If a $4 \mathrm{~kg}$ of ice is inside a closed cubical thermacol box of side length $20 \mathrm{~cm}$ and wall thickness $4 \mathrm{~cm}$ then the mass of the ice remaining after 10 hours is nearly
(The out side temperature $=50^{\circ} \mathrm{C}$ coefficient of thermal conductivity of thermacol $=0.01 \mathrm{Js}^{-1} \mathrm{~m}^{-1 \circ} \mathrm{C}^{-1}$
Latent heat of fusion of ice $=335 \times 10^{3} \mathrm{~J} \mathrm{~kg}^{-1}$ )

148215 A $10 \mathrm{H}$ inductor carries of $20 \mathrm{~A}$. If the energy stored in this inductor melts ice at $0^{\circ} \mathrm{C}$, then the mass of the ice melted is [Latent heat of ice $\left.=\mathbf{2 . 2 6} \times 10^{3} \mathrm{~J} \mathrm{~kg}^{-1}\right]$

148216 A small electric heater is used to heat $200 \mathrm{~g}$ of water. The time required to bring all this water from $40^{\circ} \mathrm{C}$ to $100^{\circ} \mathrm{C}$ is $200 \mathrm{~s}$. If specific heat of the water is $42 \mathrm{Jkg}^{-1} \mathrm{~K}^{-1}$ then the power supplied by the heater is-

148217 $5 \mathrm{~g}$ of ice at $-30{ }^{\circ} \mathrm{C}$ and $20 \mathrm{~g}$ of water at $35^{\circ} \mathrm{C}$ are mixed together in a calorimeter. The final temperature of the mixture is
(Neglect heat capacity of the calorimeter specific hear capacity of ice $=0.5 \mathrm{cal} \mathrm{g}^{-1}{ }^{0} \mathrm{C}^{-1}$ and latent heat of fusion of ice $=80 \mathrm{cal} \mathrm{g}^{-1}$ and specific hear capacity of water $=1 \mathrm{cal} \mathrm{g}^{-1}{ }^{\circ} \mathrm{C}^{-1}$ )

148218 If a $4 \mathrm{~kg}$ of ice is inside a closed cubical thermacol box of side length $20 \mathrm{~cm}$ and wall thickness $4 \mathrm{~cm}$ then the mass of the ice remaining after 10 hours is nearly
(The out side temperature $=50^{\circ} \mathrm{C}$ coefficient of thermal conductivity of thermacol $=0.01 \mathrm{Js}^{-1} \mathrm{~m}^{-1 \circ} \mathrm{C}^{-1}$
Latent heat of fusion of ice $=335 \times 10^{3} \mathrm{~J} \mathrm{~kg}^{-1}$ )

148215 A $10 \mathrm{H}$ inductor carries of $20 \mathrm{~A}$. If the energy stored in this inductor melts ice at $0^{\circ} \mathrm{C}$, then the mass of the ice melted is [Latent heat of ice $\left.=\mathbf{2 . 2 6} \times 10^{3} \mathrm{~J} \mathrm{~kg}^{-1}\right]$

148216 A small electric heater is used to heat $200 \mathrm{~g}$ of water. The time required to bring all this water from $40^{\circ} \mathrm{C}$ to $100^{\circ} \mathrm{C}$ is $200 \mathrm{~s}$. If specific heat of the water is $42 \mathrm{Jkg}^{-1} \mathrm{~K}^{-1}$ then the power supplied by the heater is-

148217 $5 \mathrm{~g}$ of ice at $-30{ }^{\circ} \mathrm{C}$ and $20 \mathrm{~g}$ of water at $35^{\circ} \mathrm{C}$ are mixed together in a calorimeter. The final temperature of the mixture is
(Neglect heat capacity of the calorimeter specific hear capacity of ice $=0.5 \mathrm{cal} \mathrm{g}^{-1}{ }^{0} \mathrm{C}^{-1}$ and latent heat of fusion of ice $=80 \mathrm{cal} \mathrm{g}^{-1}$ and specific hear capacity of water $=1 \mathrm{cal} \mathrm{g}^{-1}{ }^{\circ} \mathrm{C}^{-1}$ )

148218 If a $4 \mathrm{~kg}$ of ice is inside a closed cubical thermacol box of side length $20 \mathrm{~cm}$ and wall thickness $4 \mathrm{~cm}$ then the mass of the ice remaining after 10 hours is nearly
(The out side temperature $=50^{\circ} \mathrm{C}$ coefficient of thermal conductivity of thermacol $=0.01 \mathrm{Js}^{-1} \mathrm{~m}^{-1 \circ} \mathrm{C}^{-1}$
Latent heat of fusion of ice $=335 \times 10^{3} \mathrm{~J} \mathrm{~kg}^{-1}$ )

148215 A $10 \mathrm{H}$ inductor carries of $20 \mathrm{~A}$. If the energy stored in this inductor melts ice at $0^{\circ} \mathrm{C}$, then the mass of the ice melted is [Latent heat of ice $\left.=\mathbf{2 . 2 6} \times 10^{3} \mathrm{~J} \mathrm{~kg}^{-1}\right]$

148216 A small electric heater is used to heat $200 \mathrm{~g}$ of water. The time required to bring all this water from $40^{\circ} \mathrm{C}$ to $100^{\circ} \mathrm{C}$ is $200 \mathrm{~s}$. If specific heat of the water is $42 \mathrm{Jkg}^{-1} \mathrm{~K}^{-1}$ then the power supplied by the heater is-

148217 $5 \mathrm{~g}$ of ice at $-30{ }^{\circ} \mathrm{C}$ and $20 \mathrm{~g}$ of water at $35^{\circ} \mathrm{C}$ are mixed together in a calorimeter. The final temperature of the mixture is
(Neglect heat capacity of the calorimeter specific hear capacity of ice $=0.5 \mathrm{cal} \mathrm{g}^{-1}{ }^{0} \mathrm{C}^{-1}$ and latent heat of fusion of ice $=80 \mathrm{cal} \mathrm{g}^{-1}$ and specific hear capacity of water $=1 \mathrm{cal} \mathrm{g}^{-1}{ }^{\circ} \mathrm{C}^{-1}$ )

148218 If a $4 \mathrm{~kg}$ of ice is inside a closed cubical thermacol box of side length $20 \mathrm{~cm}$ and wall thickness $4 \mathrm{~cm}$ then the mass of the ice remaining after 10 hours is nearly
(The out side temperature $=50^{\circ} \mathrm{C}$ coefficient of thermal conductivity of thermacol $=0.01 \mathrm{Js}^{-1} \mathrm{~m}^{-1 \circ} \mathrm{C}^{-1}$
Latent heat of fusion of ice $=335 \times 10^{3} \mathrm{~J} \mathrm{~kg}^{-1}$ )