Achieve 3/4: Question #10

[Titania Le [2F]]

Hi! Can someone explain how they solved this problem?

"An ice cube with a mass of 45.8 g at 0.0 ∘C is added to a glass containing 446 g of water at 45.0 ∘C. Determine the final temperature of the system at equilibrium. The specific heat capacity of water, Cs, is 4.184 J/g⋅∘C and the standard enthalpy of fusion, ΔH∘fus, of water is 6.01×103 J/mol. Assume that no energy is transferred to or from the surroundings."

Thank you :)

[Sydney Trieu 2B]

You have to use the equation q = mCsΔT. So you set one side of the equation for the ice cube and the other side of the equation for the water and set them equal to each other. But on the side with the ice cube, you have to add the energy needed to melt the ice, which is equal to the number of moles in the ice cube multiplied by the enthalpy of fusion. Then you just need to solve for Tfinal. It should look something like this:

[(45.8)(4.184)(Tfinal - 0)] + [(45.8/18.02)(6.01 x 10^3)] = -[(446g)(4.184)(Tfinal - 45)]