how many joules are required to heat a frozen can of juice (360 grams) from -5C (the temperature of an overcooled refrigerator to 110C) the highest practical temperature within a microwave oven?

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  1. This is a long but not difficult problem.
    q1 = heat to move solid from -5 to zero C.
    q1 = mass x specific heat (I assume you will use that of ice) x 5).

    q2 = heat to melt solid.
    q2 = mass x heat fusion.

    q3 = heat to move liquid from zero to 100.
    q3 = mass x specific heat liquid x delta T.
    delta T = 100

    q4 = heat to vaporize liquid to vapor at 100.
    q4 = mass x delta Hvap.

    q5 = heat to move steam T from 100 to 110 C.
    q5 = mass x specific heat steam x 10

    total Q = q1 + q2 + q3 + q4 + q5

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  2. with sig figs: 1,000,000

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  3. So this could be completely wrong lol. But heres my guess...
    (360g) * (1.87 j/g C) * (110) = 74052 J
    make sure to thumbs down if wrong!

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