Specific heat question from Chad's video

[bretonnia]

---

Members don't see this ad.

In one of the quizzes in Chad's calorimetry lecture, he asks:
After a 50g piece of an unknown metal at 50°C is added to 50g of H2O(l) at 0°C, the mixture reaches thermal equilibrium at 10°C. What is the specific heat of the metal? (specific heat of water is 1cal/g×°C)

In the solution, he begins by stating the equation: q(water) + q(metal) = 0

I can do the rest, but I don't understand why he set q(water) + q(metal) equal to zero. Is it because the water and metal are at thermal equilibrium?

Thanks in advance.

 

[LetsGo2DSchool]

It's because the heat released by the metal is equal to the heat gained by the water.

Thus, -qmetal = qwater (same thing as qmetal = -qwater)
and why qwater + qmetal = 0

Another way to view it is this is a closed environment, so the heat released by the system (metal) is gained by the surroundings (water) and therefore, the net effect is equal to 0.

 

[bretonnia]

You're awesome . . .

 

[UCF FINatic]

One way I look at this question is that q=MC(delta)T.

So you know its a contained system. Thus, q=MC(delta)T of Water must equal q=MC(delta)T of the Metal. So just set up the equation like MC(delta)T=MC(delta)T, plug in the given for the metal on one side and the given for water on the otherside (sidenote: The Otherside is a great RHCP song!), then plug and chug baby.

 

[bretonnia]

Thanks! You're awesome too!

 

[EpicDentist]

To set up this equation, would it be the following:

Delta H = (50 g unknown metal)(1)(50-10 degrees Celsius)

Correct?

 

[woox]

Whats the answer? C(of unknown metal) = -0.25?

 

[jwnichols21]

One way I look at this question is that q=MC(delta)T.

So you know its a contained system. Thus, q=MC(delta)T of Water must equal q=MC(delta)T of the Metal. So just set up the equation like MC(delta)T=MC(delta)T, plug in the given for the metal on one side and the given for water on the otherside (sidenote: The Otherside is a great RHCP song!), then plug and chug baby.

This is wrong. I initially did it like this and ended up with -0.25, which is incorrect. The first few posts are correct in stating that you have to have them sum to 0 in order to have your signs straight; and this is due to the heat loss of one system being absorbed by the other.