thank you EletricPavlov for hemping me solve the other problem :)
Suppose a 300 g brass ball (specific heat= 0.38) at 800 degrees celsius is placed into a cup containing 50 g of ice at -25 degrees celsius. What final temperature does this system reach?
thx!!
+37146 Again, the heat the brass ball LOSES is GAINED by the ICE (s=2.05)and the WATER (s=4.188) that forms from the melting ice AND some of the heat is needed to MELT the ice (Latent heat of fusion (or melting) of ice is 334 j/gm)
Brass heat loss = 300 (.38) (800- Tf) where Tf is the final temp
Heat GAINED by ice 50 (2.05)(25) 25 is the temp change to get the ice to 0 degrees where it melts
Melting ice 50(334)
Heat gained by the WATER (after the ice melts) = 50(4.188)(Tf) (since the water warms from 0 degrees to Tf
now equate the losses and gains
300(.38)(800- Tf) = 50 (2.05)(25) + 50(334) + 50(4.188)(Tf)
Solve for Tf
91,200 - 114Tf = 2,562.5 +16,700 + 209.4Tf
71937.5 = 323.4 Tf
Tf = 222.44 degrees C
+37146 Again, the heat the brass ball LOSES is GAINED by the ICE (s=2.05)and the WATER (s=4.188) that forms from the melting ice AND some of the heat is needed to MELT the ice (Latent heat of fusion (or melting) of ice is 334 j/gm)
Brass heat loss = 300 (.38) (800- Tf) where Tf is the final temp
Heat GAINED by ice 50 (2.05)(25) 25 is the temp change to get the ice to 0 degrees where it melts
Melting ice 50(334)
Heat gained by the WATER (after the ice melts) = 50(4.188)(Tf) (since the water warms from 0 degrees to Tf
now equate the losses and gains
300(.38)(800- Tf) = 50 (2.05)(25) + 50(334) + 50(4.188)(Tf)
Solve for Tf
91,200 - 114Tf = 2,562.5 +16,700 + 209.4Tf
71937.5 = 323.4 Tf
Tf = 222.44 degrees C
+37146 Of note.....the water will have to be under pressure or some of it will turn to steam.
If steam is allowed then the max temperature reached will be 100 degrees C and some of the water will be turned to steam at 100 degrees
How much?
The latent heat of vaoprization of water is 2230 j/gm so if we are making steam the equation becomes:
300(.38)(800- 100) = 50 (2.05)(25) + 50(334) + 50(4.188)(100) + Wm(2230) where Wm = mass of water turned to steam
Wm = 17.75 gms of steam and the entire system remains at 100 C
