an ice cube initially at -10 has a mass of 9.0g and is placed on a 2.0kg plate (cp=0.5j/gk) If the temp drops by 5 what is the final temp of the H2O (cice=2j/gk, Hfus=6.00kj/mol)
the2 kg plate has lost heat = 2000g x 0.5 x 5 = 5000 joules
So ice has gained 5000 joules .
Ice heat 9.0 x 2 x10 + 333.3 + 9.0 + 9.0 temp diffx 4.18 = 5000
9.0 x temp diff.x 4.18 = 5000-3000-180
9.0 x temp. Diff. = 1820/4.18
Temp diff = 435.4/9 = 48.4 C
Answer: the final temp of water is 48.4 C
Content will be erased after question is completed.
Enter the email address associated with your account, and we will email you a link to reset your password.
Forgot your password?