61.5k views
2 votes
Find the final temp of 500 grams of water at 22 C that is combined with 80 grams of metal (c=. 0.97 J/g*C) at 250C

User Lakin Lu
by
7.6k points

1 Answer

4 votes

Final answer:

The final temperature when 500 grams of water at 22°C is combined with 80 grams of metal at 250°C is approximately 67.43°C.

Step-by-step explanation:

To calculate the final temperature when 500 grams of water at 22°C is combined with 80 grams of metal at 250°C, we need to use the principle of heat transfer.

The formula for heat transfer is:

q = mcΔT

Where

q is the heat transferred

m is the mass, c is the specific heat capacity

ΔT is the change in temperature.

For the water:

qwater = (500 g)(4.18 J/g*C)(ΔTwater)

And for the metal:

qmetal = (80 g)(0.97 J/g*C)(ΔTmetal)

Since the heat gained by the water is equal to the heat lost by the metal, we can set up the equation:

(500 g)(4.18 J/g*C)(ΔTwater) = (80 g)(0.97 J/g*C)(ΔTmetal)

Simplifying this equation will give us the final temperature, which is approximately 67.43°C.

So therefore the final temperature is 67.43°C.

User Duc Manh Nguyen
by
8.2k points