A 225-g sample of aluminum was heated to 125.5 oc, then placed into 500.0 g water at 22.5 oc. (the specific heat of aluminum is 0.900 j/goc). calculate the final temperature of the mixture. (assume no heat loss to the surroundings.)

when heat gained = heat lost

when AL is lost heat and water gain heat

∴ (M*C*ΔT)AL = (M*C*ΔT) water

when M(Al) is the mass of Al= 225g

C(Al) is the specific heat of Al = 0.9

ΔT(Al) = (125.5 - Tf)

and Mw is mass of water = 500g

Cw is the specific heat of water = 4.81

ΔT = (Tf - 22.5)

so by substitution:

∴225* 0.9 * ( 125.5 - Tf) = 500 * 4.81 * (Tf-22.5)

∴Tf = 30.5 °C

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floweregg floweregg · Answer 2 · 2020-09-28T23:35:03+02:00

Answer:

ans: the final temperature is 31.6 degrees Celsius

Explanation:

knowns:

m(alum)=225g

Δt(alum)= (125.5C-final temp)

sp.h(alum)=0.900J/gC

m(wtr)=500.0g

Δt(wtr)=(final temp- 22.5C)

sp.h(wtr)=4.184 J/gC

temp final=x

alum: h=(225g)(0.900J/gC)(125.5C-x)

water: h=(500.0g)(4.184J/gC)(x-22.5C)

heat gained by water=heat lost by metal

(225g)(0.900J/gC)(125.5C-x)=(500.0g)(4.184J/gC)(x-22.5C)

step one: mass multiplied by specific heat on both sides

(left side)225g * .900J/gC; (right side)500.0g * 4.184J/gC

202.5J/C (125.5C-x) = 2092J/C (x-22.5C)

step two: cross multiply on both sides

(left side) 202.5J/C * 125.5C and 202.5J/C * -x;

(right side) 2092J/C * x and 2092J/C * -22.5C

25413.75J - 202.5xJ/C = 2092xJ/C - 2114.5J

step three: add on both sides

(left side) 25413.75J + 2114.5J

(right side) 2092xJ/C + 202.5xJ/C

72483.75J = 2294.5xJ/C

step four: divide on the left side (everything is divided but x)

(left side) 72483.75J / 2294.5J/C

31.59021573 C = x

step five: sig figs + answer

31.6 = x

x= the final temperature is 31.6 C

I hope this helps, took me forever to figure out for homework so I figured that I might as well share how I got my answer! Feel free to tell me if I typed in anything incorrectly :)