146658
Match the following List-I with List-II
| |List-I | | List-II |
| :--- | :--- | :--- | :--- | :--- |
| A | When ice melts into water | I | Volume increase |
| B | When water changes into steam | II | Volume decrease |
| C | Melting point of ice | III | Increases with increase of pressure |
| D | Boiling point of water | IV | Decreases with increase in pressure |
146659 Water of volume 2 litre in a container is heated with a coil of $1 \mathrm{~kW}$ at $27^{\circ} \mathrm{C}$. The lid of the container is open and energy dissipates at rate of $160 \mathrm{~J} / \mathrm{s}$. In how much time temperature will rise from $27^{\circ} \mathrm{C}$ to $77^{\circ} \mathrm{C}$ ? [Given specific heat of water is $4.2 \mathrm{~kJ} / \mathrm{kg}$ ]
146660
Find the quantity of heat required to convert $40 \mathrm{gm}$ of ice at $-20^{\circ} \mathrm{C}$ into water at $20^{\circ} \mathrm{C}$. Given $\mathrm{L}_{\text {ice }}=0.336 \times 10^{6} \mathrm{~J} / \mathrm{kg}$.
Sp heat of ice $=2100 \mathrm{~J} / \mathrm{kg}-\mathrm{K}$ sp heat of water $=$ $4200 \mathrm{~J} / \mathrm{kg}-\mathrm{K}$
146658
Match the following List-I with List-II
| |List-I | | List-II |
| :--- | :--- | :--- | :--- | :--- |
| A | When ice melts into water | I | Volume increase |
| B | When water changes into steam | II | Volume decrease |
| C | Melting point of ice | III | Increases with increase of pressure |
| D | Boiling point of water | IV | Decreases with increase in pressure |
146659 Water of volume 2 litre in a container is heated with a coil of $1 \mathrm{~kW}$ at $27^{\circ} \mathrm{C}$. The lid of the container is open and energy dissipates at rate of $160 \mathrm{~J} / \mathrm{s}$. In how much time temperature will rise from $27^{\circ} \mathrm{C}$ to $77^{\circ} \mathrm{C}$ ? [Given specific heat of water is $4.2 \mathrm{~kJ} / \mathrm{kg}$ ]
146660
Find the quantity of heat required to convert $40 \mathrm{gm}$ of ice at $-20^{\circ} \mathrm{C}$ into water at $20^{\circ} \mathrm{C}$. Given $\mathrm{L}_{\text {ice }}=0.336 \times 10^{6} \mathrm{~J} / \mathrm{kg}$.
Sp heat of ice $=2100 \mathrm{~J} / \mathrm{kg}-\mathrm{K}$ sp heat of water $=$ $4200 \mathrm{~J} / \mathrm{kg}-\mathrm{K}$
146658
Match the following List-I with List-II
| |List-I | | List-II |
| :--- | :--- | :--- | :--- | :--- |
| A | When ice melts into water | I | Volume increase |
| B | When water changes into steam | II | Volume decrease |
| C | Melting point of ice | III | Increases with increase of pressure |
| D | Boiling point of water | IV | Decreases with increase in pressure |
146659 Water of volume 2 litre in a container is heated with a coil of $1 \mathrm{~kW}$ at $27^{\circ} \mathrm{C}$. The lid of the container is open and energy dissipates at rate of $160 \mathrm{~J} / \mathrm{s}$. In how much time temperature will rise from $27^{\circ} \mathrm{C}$ to $77^{\circ} \mathrm{C}$ ? [Given specific heat of water is $4.2 \mathrm{~kJ} / \mathrm{kg}$ ]
146660
Find the quantity of heat required to convert $40 \mathrm{gm}$ of ice at $-20^{\circ} \mathrm{C}$ into water at $20^{\circ} \mathrm{C}$. Given $\mathrm{L}_{\text {ice }}=0.336 \times 10^{6} \mathrm{~J} / \mathrm{kg}$.
Sp heat of ice $=2100 \mathrm{~J} / \mathrm{kg}-\mathrm{K}$ sp heat of water $=$ $4200 \mathrm{~J} / \mathrm{kg}-\mathrm{K}$
146658
Match the following List-I with List-II
| |List-I | | List-II |
| :--- | :--- | :--- | :--- | :--- |
| A | When ice melts into water | I | Volume increase |
| B | When water changes into steam | II | Volume decrease |
| C | Melting point of ice | III | Increases with increase of pressure |
| D | Boiling point of water | IV | Decreases with increase in pressure |
146659 Water of volume 2 litre in a container is heated with a coil of $1 \mathrm{~kW}$ at $27^{\circ} \mathrm{C}$. The lid of the container is open and energy dissipates at rate of $160 \mathrm{~J} / \mathrm{s}$. In how much time temperature will rise from $27^{\circ} \mathrm{C}$ to $77^{\circ} \mathrm{C}$ ? [Given specific heat of water is $4.2 \mathrm{~kJ} / \mathrm{kg}$ ]
146660
Find the quantity of heat required to convert $40 \mathrm{gm}$ of ice at $-20^{\circ} \mathrm{C}$ into water at $20^{\circ} \mathrm{C}$. Given $\mathrm{L}_{\text {ice }}=0.336 \times 10^{6} \mathrm{~J} / \mathrm{kg}$.
Sp heat of ice $=2100 \mathrm{~J} / \mathrm{kg}-\mathrm{K}$ sp heat of water $=$ $4200 \mathrm{~J} / \mathrm{kg}-\mathrm{K}$