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?

Answer 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

Answer 2

with sig figs: 1,000,000

Answer 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!