Specific Heat Question: Calculating Temperature Change with Copper and Water

[Yoruichi]

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? @@

 

[SteamKing]

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? @@

You've assumed the opposite direction of heat flows in the various components of this system.

For example, take the 150-gram piece of copper which is heated to a temperature of 80 °C and then placed in the water. It is reasonable to assume that Tf in this case is going to be less than 80 °C. Similarly, for the water which has an initial temperature of 20 °C, it would be reasonable to assume that Tf is going to be greater than the initial temperature.

You need to correct the temperature differences in your equations to reflect these facts.

 

[Yoruichi]

You've assumed the opposite direction of heat flows in the various components of this system.

For example, take the 150-gram piece of copper which is heated to a temperature of 80 °C and then placed in the water. It is reasonable to assume that Tf in this case is going to be less than 80 °C. Similarly, for the water which has an initial temperature of 20 °C, it would be reasonable to assume that Tf is going to be greater than the initial temperature.

You need to correct the temperature differences in your equations to reflect these facts.

Oh now I get it! How clumsy am I haha, thanks for you help :D