Chemistry question?
A 110.-g sample of copper (specific heat capacity = 0.20 J/°C•g) is heated to 82.4 °C and then placed in a container of water at 22.3 °C. The final temperature of the water and copper is 26.0 °C. What is the mass of the water in the container, assuming that all heat lost by the copper is gained by the water?
___________g H2O
Hint: Set heat gained by water equal to heat lost by copper and solve for mass of water. Keep all quantities positive to avoid sign errors.
Answer:
From the formula Q = mc*deltaT
copper
m = 110 g
c = 0.2
delta T = 82.4 - 26
water
m = X g
c = 4.2 (this is the standard value you should know)
delta T = 26 - 22.3
Qcopper = Q water
110*0.2*(82.4-26) = X*4.2*(26-22.3)
X = 79.85 g
when you guys post science questions you should post the formulas with them.
If you had, it would be a quick 2 minutes to answer, but to look up the formulas and then do the math is too time consuming. post the formulas relating mass/mole number to heat capacity and it is a quick computation.
No more science books on my desk
The answers post by the user, for information only, FunQA.com does not guarantee the right.
More Questions and Answers: