Specific heat question help?

[匿名]

2017-07-19 8:29 am

A 225-g sample of aluminum was heated to 125.5C, then placed into 500.0 g water at 22.5C (the specific heat of aluminum is .900J/gC). Calculate the final temperature of the mixture. (Assume no heat loss to the surrounding).

I understand how to set the formula up

(225.0g)(.900j/gC)(125.5C-x)=(500)(4.184)(x-22.5)

Please explain in full detail what to do from here on out. Thank you

更新1:

sorry (500)(4.184)(x-22.5)

回答 (1)

不用客氣

2017-07-19 8:43 am

Let x°C be the final temperature.

For the sample of aluminum :
Mass, m = 225.0 g
Specific heat, c = 900 J/g°C
Temperature drop, ΔT = (125.5 - x)°C
Hence, energy lost by aluminum = m c ΔT = 225.0 × 900 × (125.5 - x) J

For the water :
Mass, m = 500.0 g
Specific heat, c = 4.184 J/g°C
Temperature raise, ΔT = (x - 22.5)°C
Hence, energy gained by water = m c ΔT = 500.0 × 4.184 × (x - 22.5) J

There is no heat loss to the surroundings. Then,
Energy lost by aluminum = Energy gained by water
225.0 × 900 × (125.5 - x) = 500.0 × 4.184 × (x - 22.5)

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