Question

A 3.00-g sample of aluminum pellets (specific heat capacity=0.89 J/°C g) and a 18.50-g sample of iron pellets (specific heat capacity = 0.45 J/°C-g) are heated to 100.0 °C. The mixture of hot iron and aluminum is then dropped into 77.4 g water at 22.0 °C. Calculate the final temperature of the metal and water mixture, assuming no heat loss to the surroundings. Final temperature = 20.23 °C An error has been detected in your answer. Check for typos. miscalculations etc. before submitting your answer. Submit Answer Try Another Version 5 item attempts remaining

Answer 1

We can use the following formula

Q = mc∆T

Q = heat energy (Joules, J), m = mass of a substance (g)

c = specific heat (units J/g∙^oC), ∆ is a symbol meaning "the change in"

∆T = change in temperature (^oC Celcius)

Heat lost by aluminium + Heat lost by Iron = Heat gained by water

For Aluminium

m = 3 g c = 0.89 J/g∙^oC   Ti = 100 ^oC Tf = ?

For Iron

m = 18.5 g c = 0.45 J/g∙^oC   Ti = 100 ^oC Tf = ?

For water

m = 77.4 g c = 4.184 J/g∙^oC   Ti = 22 ^oC Tf = ?

Let us do calculation

3 g x 0.89 J/g∙^oC x (100 ^oC - Tf ) + 18.5 g x 0.45 J/g∙^oC x (100 ^oC - Tf ) =  77.4 g x 4.184 J/g∙^oC x (Tf - 22 ^oC)

(267 - 2.67 Tf) + (832.5 - 8.325 Tf) = 323.84 Tf - 7,124.51

334.835 Tf = 8224.0152

Tf = 24.56 ^oC

FInal temperature of the water and MIXTURE  is 24.56 ^oC