1 kg of ice at 0° C is mixed with 9 kg of water at 50° C (The latent heat of ice is 3.34x105 J/kg and the specific heat capacity of water is 4160 J/kg). What is the resulting temperature?

Answer 1

Heat into ice = 1*3.34*10^5 +(T-0)(4160)
Heat out of water = 4160 (50-T)(9)

set equal and solve for T

Answer 2

omg thank you so much!!

Answer 3

but ? where did the 1.3 come frm

Answer 4

1*3 is one times 3
1 kilogram times latent heat of ice

Answer 5

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Answer 6

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