Determine the final temp when 45.8g of aluminum at -5.2C is added to a mixture of 45.0g of ice at 0C and 2000g of water at 95C?

Answer 1

[45.8 g Al x specific heat Al x (Tf -Ti)]+ [45 g ice x heat fusion ice]+[45 g H2O from melted ice x specific heat water x (Tf - Ti)] + [2000 g H2O x specific heat water x (Tf-Ti)] = 0
Tf = final T; Ti = initial T. Solve for Tf. Post your work if you get stuck.

Answer 2

I know the specific heat for water is 4.184 and Al is 0.900. I don't know ice - our professor never went over that. Is it the same as water, or different?

Answer 3

No it isn't; however, you don't need the specific heat of ice since the temperature of the ice is listed as zero C. What you need is the heat of fusion (to melt the ice) and that is 334 J/g.

Answer 4

che che che chiia