Textbook 4C.13

[Anna Turk 1D]

An ice cube of mass 50.0 g at 0°C is added to a glass containing 400.0 g of water at 45°C. What is the final temperature of the system (see Tables 4A.2 and 4C.1)? Assume that no heat is lost to the surroundings.
I'm having trouble solving this, does anyone know the steps to complete this problem? Thanks!

[Cadence Chang]

Hi there!

When the ice is added to the liquid water, it undergoes a phase change which releases energy. This released energy is then absorbed by the liquid water. Therefore, you want to set the energy released (by the ice) equal to the negative amount of the energy absorbed (by the liquid).

q_released = – q_absorbed

The energy released by the ice is equal to the energy from the phase change: mol*∆Hfusion + gCsp∆T. ∆T = Tf - 0, since Tf is what we are solving for and 0 is the initial temperature of the water.

The energy absorbed by the liquid water (since it's not undergoing a phase change) is just gCsp∆T where ∆T = Tf - 45. Then you just set the equations equal to each other and solve. Make sure you remember the negative sign in the equation above!

Hope this helps!

[Sophia Schiro]

The energy lost to heat by the ice equals the energy gained by the water, so you can set them equal to each other. This would be n*enthalpy of fusion + mc(Tf-0) = mc(Tf-45), which you can solve for T.