Find out the amount of heat required to convert 1 gm of ice at -30 ⁰C to 1 gm of steam at 120
⁰C. Given the specific heat of ice is 2090 J/kg/⁰C, specific heat of water 4180 J/kg/⁰C, specific heat
of steam is 2010 J/kg/⁰C, latent heat of ice to water is 333000 J/kg and latent heat of water to steam
is 2260000 J/kg.
Plz solve it....

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  1. You gave me all the info you need.
    Do it for one KILOGRAM because those are your units. DIVIDE answer by 1000 then.

    Heat the ice from -30 to 0
    30 * 2090 J

    melt the ice
    333000 J

    heat the resulting water 100 degrees
    100 * 4180 Joules

    boil the water
    2,260,000 Joules

    heat the steam
    (120-100)(2,010) Joules

    Add them up and divide by 1000 to get a gram instead of a kilogram

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