Question

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

This is a chemistry question. You have to know the equation for the specific heat which i believe equates to enthalpy

Answer 2

This is what I said

Q1 + Q2= Q3
mi* L + mi * c * (x - 0) = mw * c * (50 - x)
48.1 * 3.34x10^5 + 48.1 * 4160 * x = 9 * 4160 ( 50 - x)
1.61 x 10^7 + 2.0 x 10^5 x = 1.87 x 10^6 - 3.74 x 10^4 x
2.0 x 10^5 x + 3.74 x 10^4 x = 1.87 x 10^6 - 1.61 x 10^7
x = -1.42 x 10^7/2.37 * 10^5 = -59.9
The resulting temperature is negative meaning the resulting temperature is 0.

@Johnbc

Answer 3

I am not 100% sure you will have to carry this over to the Chemistry section of the website.