If 100. g of iron at 100.0 C is placed in 200. g of water at 20.0 C in an insulated container, what will the temp, C, of the iron and the water ( C value is 4.18) when both are at the same temp?

The specific heat of iron is 0.444 KJ/Kg C

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steps would be appreciated... thanks!!!

If 100. g of iron at 100.0 C is placed in 200. g of water at 20.0 C in an insulated container, what will the temp, C, of the iron and the water ( C value is 4.18) when both are at the same temp?

The easy way to work these is to write an equation that the sum of the heats gained is zero (one will lose heat, or negative heat gained).

Heatgainedwater + heatgainediron=0
mw*cw*(Tf-Tiw)+ mi*Ci*(Tf-Tii)=0
you know everything except Tf, and you solve for that.

thanks!!!!

To solve for the final temperature (Tf), we can use the equation:

mw * cw * (Tf - Tiw) + mi * Ci * (Tf - Tii) = 0

where:
- mw is the mass of water (200 g)
- cw is the specific heat capacity of water (4.18 J/g °C)
- Tf is the final temperature of both substances
- Tiw is the initial temperature of water (20.0 °C)
- mi is the mass of iron (100 g)
- Ci is the specific heat capacity of iron (0.444 J/g °C)
- Tii is the initial temperature of iron (100.0 °C)

Plugging in the known values, we get:

200g * (4.18 J/g °C) * (Tf - 20.0 °C) + 100g * (0.444 J/g °C) * (Tf - 100.0 °C) = 0

Simplifying the equation gives:

836(Tf - 20) + 44.4(Tf - 100) = 0

836Tf - 16720 + 44.4Tf - 4440 = 0

880.4Tf - 21160 = 0

880.4Tf = 21160

Tf = 21160 / 880.4

Tf ≈ 24.07 °C

Therefore, when both substances reach the same temperature, it will be approximately 24.07 °C.