- #1
sadifermi
- 5
- 0
I need help with a question. I think I have an answer, but don't know if it's right. It also doesn't add up to 0 if I plug in the final temperture into the equation. But I keep getting the same answer. Help please!
A 150 g piece of copper at 90°C is placed in 250 g of water at 20°C. Assume that no heat escapes into the surroundings. Find the final temperature of the mixture.
mass of copper: (0.150kg)
specific heat capacity: (390J/(kg°C))
temperture of copper: (t2-90°C)
mass of water: (0.250kg)
specific heat capacity: (4200J/(kg°C))
temperture of water: (t2-20°C))
equation: (mass of copper)(spec. heat cap.)(final temp - intial temp) + (mass of water)(spec. heat cap.)(final temp - initial temp) = 0
(0.150kg)(390J/(kg°C))(t2-90°C)+(0.250kg)(4200J/(kg°C))(t2-20°C))=0
1108.5t2=26265
t2 = 23.7°C
A 150 g piece of copper at 90°C is placed in 250 g of water at 20°C. Assume that no heat escapes into the surroundings. Find the final temperature of the mixture.
mass of copper: (0.150kg)
specific heat capacity: (390J/(kg°C))
temperture of copper: (t2-90°C)
mass of water: (0.250kg)
specific heat capacity: (4200J/(kg°C))
temperture of water: (t2-20°C))
equation: (mass of copper)(spec. heat cap.)(final temp - intial temp) + (mass of water)(spec. heat cap.)(final temp - initial temp) = 0
(0.150kg)(390J/(kg°C))(t2-90°C)+(0.250kg)(4200J/(kg°C))(t2-20°C))=0
1108.5t2=26265
t2 = 23.7°C
