Question

A 3.00-g sample of aluminum pellets (specific heat capacity = 0.89 J/°C·g) and a 11.00-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 73.8 g water at 22.0 °C. Calculate the final temperature of the metal and water mixture, assuming no heat loss to the surroundings. Please be super specific on how you get to each step!

Answer 1

Given,

Mass of sample of Al pellets = 3.00 g

Specific heat capacity of the aluminum pellets (C_Al) = 0.89 J/g ^oC

Mass of sample of iron pellets = 11.00 g

Specific heat capacity of the iron pellets (C_Fe) = 0.45 J/g ^oC

The initial temperatures of Al pellets and Fe pellets = 100.0 ^oC

Mass of water = 73.8 g

The initial temperature of water = 22 ^oC

Also, We know,

Specific heat capacity of water = 4.18 J /g ^oC

Now, When the mixture of hot iron and aluminum is dropped into water,

Heat loss by the mixture of hot iron and aluminum = Heat gained by the water

- [ (m x C_Al x T) + (m x C_Fe x T) ] = m x C x T

Suppose the final temperature of the metal and water mixture as "c"

Substituting the known values in the expression,

- [ (3.00 g x 0.89 J/g ^oC x (c - 100.0) ^oC) + (11.00 g x 0.45 J/g ^oC x (c - 100.0) ^oC) ] = 73.8 g x 4.18 J/g ^oC x (c - 22) ^oC

- [ 2.67 c - 267 + 4.95 c - 495 ] = 308.484 c - 6786.6

-2.67 c + 267 - 4.95 c + 495 ] = 308.484 c - 6786.6

-316.104 c = -7548.65

c = 23.88 ^oC

Thus, the final temperature of metal and water mixture is 23.8 ^oC Or 23.9 ^oC.