- #1
Yoruichi
- 17
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Homework Statement
A copper container (mass: 100 g) holds 200 g of water. The temperature of the container and water is 20°C.
A 150-g piece of copper heated to 80°C is placed in the water. The water is stirred thoroughly. After sufficient time elapses, to what temperature does the water change? From 1 - 5 below choose the best answer. Assume that no heat is transferred to or from the environment. The specific heat of water is 4.2J/g.K, and the specific heat of copper is 0.4 J/g.K.
Homework Equations
Q=mc delta t
The Attempt at a Solution
My attempt is that since no heat is transfer to or from the environment, the heat lose by the 150g piece of copper is equal to the the heat gain by the copper container and the water in it.
Let the final temperature be Tf
Let Q1 be the heat energy lose by the 150-g piece of copper
Q = mc delta T
Q1 = 150g x 0.4J/g.k x (Tf - 80°C)
= 60 Tf - 4800
Let Q2 be the heat energy gained by the copper container and the water in it
Q2 = {100g x 0.4J/g.k x (Tf - 20°C)} + {200g x 4.2J/g.k x (Tf - 20°C)}
= 40 Tf - 800 + 840 Tf - 16800
= 880 Tf - 17600
Q1 = Q2
60 Tf - 4800 = 880 Tf - 17600
820 Tf = 12800
Tf = 15.61°C
The answer is wrong, may I know what's the problem of my answer? @@