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A colorimeter contains 400 g of water at...

A colorimeter contains `400 g` of water at a temperature of `5^(@)C`. Then, `200 g` of water at a temperature of `+10^(@)C` and `400 g` of ice at a temperature of `-60^(@)C` are added. What is the final temperature of the contents of calorimeter?
Specific heat capacity of water `-1000` cal`//kg//K`
Specific latent heat of fusion of ice `=80xx1000` cal`//kg`
Relative specific heat of ice `=0.5`

Text Solution

Verified by Experts

Let t be the common temperature above zero celsius.
Heat lost by calorimeter and water added
`=400xx10^(-3)xx1000(5-t)+200xx10^(-3)xx1000(10-t)`
Heat gained by ice `=400xx10^(-3)xx5000(60+t)+400xx10^(-3)xx80000+400xx10^(-3)xx1000t`
Since Heat lost = heat gained `therefore400(5-t)+200(10-t)=200(60+t)+400xx80+400t`
or `t=-24.33^(@)C`
This is contradictory of the assumption that `t` is above zero.
Hence, the final temperature is either `0^(@)C` or less than `0^(@)C. Let the temperature be `t^(*@)C` (less than `0^(@)C).
Then `400xx10^(-3)xx1000xx5+200xx10^(-3)xx100xx10+600xx10^(-3)80000+600xx10^(-3)xx500xxt=400xx500(60-t)`
or `400xx5+200xx10+600xx80+300t=200(60-t)`
or `20+20+480+3t=120-2t`
or `t=-80^(@)C`
This is absurd because the lowest temperature is `-60^(@)C` and we dicard this second asumption also.
So the final temperature is `0^(@)C`.
Let x grams of water be converted into ice.
Heat lost `=400xx10^(-3)xx100xx5+200xx10^(-3)xx1000xx10+x xx10^(-3)xx8000-4000+80x`
Heat gained `=400xx10^(-3)xx500xx60`
or `4000+80x=12000`
or `x=100g`
Hence, the final result is `500g` of ice and `500g` of water at `0^(@)C`.
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A calorimeter contains 400 gm of water at temperature of 5^@C . Then, 200 gm of water at a temperature of +10^@C and 400 gm of ice at temp. of -60^@C are added. Specific heat of ice =0.5 cal/gm. At equilibrium

What will be the final temperature when 150 g of ice at 0^@C is mixed with 300 g of water at 50^@C . Specific heat of water =1 cal//g//^@C . Latent heat of fusion of ice =80 cal//g .

A piece of ice of mass 50 g exists at a temperature of -20^(@)C . Determine the total heart required to covert it completely to steam at 100^(@)C . (Specific heat capacity of ice = 0.5 cal//g-^(@)C , specific latent heat of fusion for ice = 80 cal//g and specific latent heat of vaporization for water = 540 cal//g ).

5g of water at 30^@C and 5 g of ice at -29^@C are mixed together in a calorimeter. Find the final temperature of mixture. Water equivalent of calorimeter is negligible, specific heat of ice = 0.5 cal//g-^@C and latent heat of ice =80 cal//g .

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Knowledge Check

  • What will be the final temperature when 150 g of ice at 0^@C is mixed with 300 g of water at 50^@C . Specific heat of water =1 cal//g//^@C . Latent heat of fusion of ice =80 cal//g .

    A
    `7.6^@C`
    B
    `6.7^@C`
    C
    `5.8^@C`
    D
    `8.5^@C`

  • 1g of ice at 0^(@)C is added to 5g of water at 10^(@)C . If the latent heat is 80cal/g, the final temperature of the mixture is

    A
    `5^(@)C`
    B
    `0^(@)C`
    C
    `-5^(@)C`
    D
    None of these

  • A piece of ice of mass 50 g exists at a temperature of -20^(@)C . Determine the total heart required to covert it completely to steam at 100^(@)C . (Specific heat capacity of ice = 0.5 cal//g-^(@)C , specific latent heat of fusion for ice = 80 cal//g and specific latent heat of vaporization for water = 540 cal//g ).

    A
    `26500 cal`
    B
    `12400 cal`
    C
    `36500 cal`
    D
    `46500 cal`