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11 April, 01:26

Compute the quantity of heat released by 25.0 g of steam initially at 100.0oC, when it is cooled to 34.0°C and by 25.0 g of water initially at 100.0 oC, when it is cooled to 34.0°C.

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  1. 11 April, 01:52
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    For steam, heat released E = 15.26KJ

    For water, heat released E2 = 6.91KJ

    Explanation:

    Given;

    Mass (steam) ms = 25g

    Mass (water) mw = 25g

    Change in temperature of both steam and water ∆T = 100-34 = 66°C

    Specific heat of water C = 4.186 J/g.°C

    Specific Latent heat L = 334J/g

    For steam;

    Heat released E = msL + msC∆T

    E = (25*334) + (25*4.186*66)

    E = 15256.9J

    E = 15.26KJ

    For water;

    Heat released E2 = mwC∆T

    E2 = 25*4.186*66

    E2 = 6906.9J

    E2 = 6.91KJ
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