A 3.10-g sample of aluminum pellets (specific heat capacity = 0.89 ) and a 10.90

Question

A 3.10-g sample of aluminum pellets (specific heat capacity = 0.89 ) and a 10.90-g sample of iron pellets (specific heat capacity = 0.45 ) are heated to 111.0°C. The mixture of hot iron and aluminum is then dropped into 92.6 g of water at 16.0°C. Calculate the final temperature of the metal and water mixture, assuming no heat loss to the surroundings. The specific heat capacity of water is 4.18 . * have gotten this far but sadly do not have the math skills to finish. PLease please show every step of the math casue I have a gap in my education as far as solving mathmatically. (3.10g)(.89)(111.0 C-T final) + (10.9g)(.45) (111.0 C-T final)= (92.6g)(4.18)(T initial-16.0 C) Thank you in advance!

Explanation / Answer

Heat lost by metals = heat gained by water

let Tf be the final temperature

3.1 x 0.89 x (111 - Tf) + 10.9 x 0.45 x (111 - Tf) = 92.6 x 4.18 x (Tf - 16)

306.25 - 2.76Tf + 544.455 - 4.905Tf = 387.068Tf - 6193.088

394.733Tf = 7043.793

Tf = 17.844 oC

So final temperature = 17.844 oC

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