1. The problem statement, all variables and given/known data If 10-kg of ice at 0 degrees Celsius is added to 2-kg of steam at 100 degrees Celsius, the temperature of the resulting mixture is? Use Joules. ANSWER: 40 2. Relevant equations Specific Heats: Ice = 2060 Steam = 2020 Latent Heats: Water melting = 3.33x10^5 Water boiling = 2.26x10^6 3. The attempt at a solution Q1=mL Q1=10(3.34x10^5) Q1=3.34x10^6 Q2=(2)(2.26x10^6) Q2=4.52x10^6 Q2-Q1=1.18x10^6 Q=MCΔT 1.18x10^6 = (2+10)(4180)(T-0) 1.18x10^6 = 5.0x10^4T T=23.6 Sadly 23.6 =/= 40 I've been trying to do this problem for over an hour now using every method I remember, stumped.