A 248-g piece of copper initially at 314 °C is dropped into 390 mL of water init

Question

A 248-g piece of copper initially at 314 °C is dropped into 390 mL of water initially at 22.6 °C. Assuming that all heat transfer occurs between the copper and the water, calculate the final temperature.

The formula is:

- (specific heat X Mass X Temp Change) for copper = (specific heat X Mass X Temp Change) for water

- (0.385 J/gC)(248g)(final temp-314 C) = (4.184 J/gC)(390g)(final temp-22.6 C)

How should I isolate and solve for the Final temperature? It's on both sides, that is what's confusing. I understand everything else.

Explanation / Answer

m(water) = 390.0 g
T(water) = 22.6 oC
C(water) = 4.184 J/goC
m(Copper) = 248.0 g
T(Copper) = 314.0 oC
C(Copper) = 0.385 J/goC
T = 25.6 oC
We will be using heat conservation equation

Let the final temperature be T oC
use:
heat lost by Copper = heat gained by water
m(Copper)*C(Copper)*(T(Copper)-T) = m(water)*C(water)*(T-T(water))
248*0.385*(314-T) = 390*4.184*(T-22.6)
95.48*(314.0-T) = 1631.76*(T-22.6)
29980.72 - 95.48*T = 1631.76*T - 36877.776
T= 38.7083 oC
Answer: 38.7 oC

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.