Verified answer

energy to cool the lead E1 = 130•300(100–T)

energy to heat water E2 = 4182•500(T–20)

energy to heat aluminum E3 = 915•200(T–20)

E1 = E2 + E3

130•300(100–T) = (T–20)(4182•500 + 915•200)

130•3(100–T) = (T–20)(4182•5 + 915•2)

390(100–T) = (T–20)(20910 + 1830)

390(100–T) = (T–20)(22740)

39000 – 390T = 22740T – 454800

23130T = 493800

T = 21.35 ºC

you can convert to K if you want.

let the final temperature is t*C

=>Q(loss) by lead = Q(gain) by water and aluminum

=>m(l)s(l)∆t = m(w)s(w)∆t + m(a)s(a)∆t

=>300 x 130 x (100-t) = 500 x 4181.6 x (t-20) + 200 x 915 x (t-20)

=>3900000 - 39000t = 2273800t - 45476000

=>t = 21.35*C

Solution:

Mass of the lead, m1 = 300 Kg

Sp. Heat of lead, S1 = 130 J/Kg-K

Mass of the Aluminum, m2 = 200 Kg

Sp. Heat of Aluminum, S2 = 915 J/Kg-K

Mass of the water, m3 = 500 Kg

Sp. Heat of water, S3 = 4200 J/Kg-K

Temperature of lead, θ1 = 100ºC

Temperature of the mixture, θm = ?

Temperature of the calorimeter, θ2 = 20ºC

Heat lost by the lead, Q1 = m1 S1 (θ1-θm)

Heat absorbed by Water & Calorimeter, Q2 = m2 S2 (θm – θ2) + m3 S3 (θm – θ2)

According to the fundamental principle of calorimetry,

Heat Lost = Heat gained

Or, Q1 = Q2

Or, m1 S1 (θ1-θm) = m2 S2 (θm – θ2) + m3 S3 (θm – θ2)

Or, m1 S1 (θ1-θm) = (θm – θ2) (m2 S2 + m3 S3)

Or, (θ1-θm) / (θm – θ2) = (m2 S2 + m3 S3) / m1 S1

= {(200 Kg X 915 J/Kg-K) + (500 Kg X 4200 J/Kg-K)} / (300 Kg X 130 J/kg-K)

= 761 / 13

Or, 13θ1 – 13θm = 761θm – 761θ2

Or, 774 θm = 13θ1 + 761θ2

= (13 X 100) + (761 X 20)

= 16520

Or, θm = 16520 / 774

Therefore, temperature of the mixture, θm = 21.34ºC