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If there is no heat loss, the heat relea...

If there is no heat loss, the heat released by the condensation of x gram of steam at `100^@C` into water at `100^@C` can be used to convert y gram of ice at `0^@C` into water at `100^@C`. Then the ratio of y:x is nearly [Given `L_l = 80 cal//gm` and `L_v= 540 cal//gm`]

A

`1:1`

B

`2:1`

C

`3:1`

D

`2.5:1`

Text Solution

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The correct Answer is:
To solve the problem, we need to equate the heat lost by the condensation of steam to the heat gained by the melting of ice and heating the resulting water. Let's break it down step by step. ### Step 1: Identify the heat loss from steam condensation When steam condenses into water at 100°C, it releases heat. The heat released by the condensation of \( x \) grams of steam can be calculated using the latent heat of vaporization (\( L_v \)): \[ \text{Heat lost} = x \cdot L_v = x \cdot 540 \, \text{cal} \] ### Step 2: Identify the heat gain by ice The heat gained by \( y \) grams of ice at 0°C to convert it into water at 100°C consists of two parts: 1. The heat required to melt the ice at 0°C into water at 0°C. 2. The heat required to raise the temperature of the resulting water from 0°C to 100°C. The heat required to melt the ice is given by: \[ \text{Heat to melt ice} = y \cdot L_l = y \cdot 80 \, \text{cal} \] The heat required to raise the temperature of the water from 0°C to 100°C is given by: \[ \text{Heat to raise temperature} = y \cdot s \cdot \Delta T = y \cdot 1 \cdot (100 - 0) = y \cdot 100 \, \text{cal} \] ### Step 3: Combine the heat gain expressions Now, we can express the total heat gained by the ice: \[ \text{Total heat gained} = \text{Heat to melt ice} + \text{Heat to raise temperature} = y \cdot 80 + y \cdot 100 = y \cdot (80 + 100) = y \cdot 180 \, \text{cal} \] ### Step 4: Set heat lost equal to heat gained According to the principle of conservation of energy (no heat loss), we can set the heat lost equal to the heat gained: \[ x \cdot 540 = y \cdot 180 \] ### Step 5: Solve for the ratio \( \frac{y}{x} \) Rearranging the equation gives us: \[ \frac{y}{x} = \frac{540}{180} = 3 \] ### Conclusion Thus, the ratio \( \frac{y}{x} \) is nearly \( 3:1 \).

To solve the problem, we need to equate the heat lost by the condensation of steam to the heat gained by the melting of ice and heating the resulting water. Let's break it down step by step. ### Step 1: Identify the heat loss from steam condensation When steam condenses into water at 100°C, it releases heat. The heat released by the condensation of \( x \) grams of steam can be calculated using the latent heat of vaporization (\( L_v \)): \[ \text{Heat lost} = x \cdot L_v = x \cdot 540 \, \text{cal} \] ...
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Knowledge Check

  • Quantity of heat lost in condensation of 10 gm of steam at 100^@ C is

    A
    `2.26 xx 10^(5) J`
    B
    `2.26 xx 10^(4) J`
    C
    `22.6 J`
    D
    `44.52 xx 10^(4) J`
  • Calculate the amount of heat (in calories) required to convert 5 gm of ice at 0^(@)C to steam at 100^(@)C

    A
    `3100`
    B
    `3200`
    C
    `3600`
    D
    `4200`
  • When x grams of steam at 100^(@)C is mixed with y grams of ice at 0^(@)C , we obtain (x + y) grams of water at 100^(@)C . What is the ratio y//x ?

    A
    1
    B
    2
    C
    3
    D
    4
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