Finding the Final Temperature of Hot Coffee in a Cold Cup

[rlc]

Homework Statement

(c11p34) You pour 160.0 g of hot coffee at 75.0 oC into a 230.0- g glass cup at 24.0 oC. If they come to thermal equilibrium quickly, what is the final temperature (in oC, enter deg in asnwer box)? Assume no heat is lost to the surroundings.

Homework Equations

(I don't even know, I've seen so many.)

The Attempt at a Solution

I've attempted this problem a couple times with different formulas, but none have worked.
Can someone just tell me what equation to use, please?

 

[BvU]

Suppose the final temperature is T_f. Then the coffee has cooled off to that temperature and the cup has heated up. They tell you no heat is lost, so apparently it's to do with heat. Where did the heat from the coffee go and where did the heat for the cup come from ?

 

[rlc]

Thank you for replying! I actually tried another equation and found the correct answer:
(mass coffee)(Temp coffee-Tf)(4.18 J/gC)=(mass glass)(Tf-temp glass)(0.840 J/gC)
(160)(75-Tf)(4.18)=(230)(Tf-24)(0.840)
Tf=63.57 deg

 

[BvU]

It's what we call a heat balance. Heat given off = heat taken up. And you did well to include the c_p.

 

[HalfLostAndSearching]

The equation you will need to use is the heat transfer equation:

Q = mcΔT

Where Q is the heat transferred, m is the mass of the substance, c is the specific heat capacity, and ΔT is the change in temperature.

In this problem, the heat transferred from the hot coffee to the cold cup will be equal to the heat gained by the cup. Therefore, we can set up the following equation:

(mcoffee)(ccoffee)(Tf - Thot) = (mcup)(ccup)(Tf - Tcold)

Where mcoffee and ccoffee are the mass and specific heat capacity of the coffee, Thot is the initial temperature of the coffee, mcup and ccup are the mass and specific heat capacity of the cup, and Tcold is the initial temperature of the cup.

We can rearrange this equation to solve for Tf, the final temperature:

Tf = (mcoffee * ccoffee * Thot + mcup * ccup * Tcold) / (mcoffee * ccoffee + mcup * ccup)

Plugging in the given values, we get:

Tf = (160.0 g * 4.18 J/g°C * 75.0°C + 230.0 g * 0.84 J/g°C * 24.0°C) / (160.0 g * 4.18 J/g°C + 230.0 g * 0.84 J/g°C)

Tf = 43.9°C

Therefore, the final temperature of the coffee in the cold cup will be 43.9°C.