[sassychar]

I need help on one question:


A 50.0g sample of copper is heated to 96.2C and then placed in a calorimeter containing 75.0g of water at 19.6C. The specific heat capacity of copper is 0.20Jg/k and that of water is 4.18J g/k. Calculate the final temperature of the metal and water assuming that all of the heat is lost by the copper is gained by the water.

I came up with this (50.0)(0.20Jg/k)(x-96.2)=(74.0)(4.18Jg/k)(x-19.6)

but the equation doesnt' seem to work. someone help me please.

[Sev]

I came up with this (50.0)(0.20Jg/k)(x-96.2)=(74.0)(4.18Jg/k)(x-19.6)

74g?

[sassychar]

I came up with this (50.0)(0.20Jg/k)(x-96.2)=(75.0)(4.18Jg/k)(x-19.6)

74g?

i meant 75.

[Sev]

I came up with this (50.0)(0.20Jg/k)(x-96.2)=(74.0)(4.18Jg/k)(x-19.6)

For delta T of copper, try using the positive value (i.e. 96.2-x).

[sassychar]

i tried that but the answer is incorrect.

the answer is suppose to be 26 C...

is the method even right?

[Borek]

Method is OK (once you correct the delta T sign problem), but it seems to me like final temp is close to 22, not 26.