so there're:(adsbygoogle = window.adsbygoogle || []).push({});

0.041 kg ice cube at 0.0 *C

0.110 kg water at 40.0 *C

in

0.062 kg iron cup at 40.0 *C

Find the equilibrium temperature of the cup and its contents:

-------

So what I tried doing was... I found the heat lost by water if cooled to 0 *C, which was 18418.4 J (mass of water * water specific heat * temperature difference which is 40)

then found th eheat needed to melt ice, which was 13735 J (mass of ice * Lf)

then found the difference of the two to find the amount of heat left... and used that much heat to warm 0.151 kg of water (mass of initial water + initial ice) at 0 *C to its final temperature... so I did, delta T=Q/(mc) which gave T=7.4 *C...

so then I used that temperature in kelvin, which is 280.6 K and did:

(specific heat of water)(mass of water which is now 0.151)(280.6+T)=(specific heat of iron)(mass of cup)(313-T)

and solved for T.... where did I make the mistake because this answer's not correct?

**Physics Forums - The Fusion of Science and Community**

# Equilibrium temperature of ice, water, and iron cup

Know someone interested in this topic? Share a link to this question via email,
Google+,
Twitter, or
Facebook

- Similar discussions for: Equilibrium temperature of ice, water, and iron cup

Loading...

**Physics Forums - The Fusion of Science and Community**