Solving Calorimetry Problem: Iron in Water

[xwhoaaax3]

okay so i understand q=mc∆T and all but this problem really has me stumped

a 12g piece of iron (80°C) has been dropped into 600g of water (25°C)

the specific heat of water is 4.184J/g°C and iron's is .44J/g°C

what is the final temperature of the two?

 

[hage567]

So you need to apply that equation for each substance. The heat lost by the iron will equal the heat gained by the water. They will both come to the same final temperature.

Try to set it up.

 

[Blueshift5]

Based on the information provided, we can use the formula q=mc∆T to solve this calorimetry problem. First, we need to calculate the heat absorbed by the water and the iron separately.

For the water, we have m=600g, c=4.184J/g°C, and ∆T=final temperature - initial temperature = final temperature - 25°C. Therefore, qwater= (600g)(4.184J/g°C)(final temperature - 25°C).

For the iron, we have m=12g, c=0.44J/g°C, and ∆T=final temperature - initial temperature = final temperature - 80°C. Therefore, qiron= (12g)(0.44J/g°C)(final temperature - 80°C).

Since the total heat absorbed by the water and the iron must be equal, we can set up the equation qwater=qiron. This gives us (600g)(4.184J/g°C)(final temperature - 25°C) = (12g)(0.44J/g°C)(final temperature - 80°C).

Solving for the final temperature, we get a value of 27.3°C. Therefore, the final temperature of the water and the iron will be approximately 27.3°C.

It is important to note that this calculation assumes that there is no heat loss to the surroundings during the process. In real experiments, it is important to account for any heat loss or gain to ensure accurate results. Additionally, the specific heat values used may vary slightly depending on the source, so it is important to make sure the correct values are used for accurate calculations.