[mc∆T]Cu + [mc∆T]HOH = 0
[mc(Tf - Ti)]Cu + [mc(Tf - Ti)]HOH = 0
All data is given in problem except T-final. Solve for Tf.
m(Cu) = 33.2 gm
m(HOH)= 50.0 gm
Ti(Cu) = 70.5 C
Ti(HOH)= 25.0 C
Tf = ???
c(Cu) = 0.385 J/gC
c(HOH)= 4.184 J/gC
Your answer should be between the temp extremes 25C & 70.5C.
A hot lump of 33.2 g of copper at an initial temperature of 70.5 °C is placed in 50.0 mL of H2O initially at 25.0 °C and allowed to reach thermal equilibrium. What is the final temperature of the copper and water given that the specific heat of copper is 0.385 J/(g·°C)? Assume no heat is lost to surroundings.
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