Question

A student has calibrated his/her calorimeter and finds the heat capacity to be 15.6 J/°C. S/he then determines the molar heat capacity of aluminum. The data are: 24.1 g Al at 100.0°C are put into the calorimeter, which contains 99.6 g H2O at 19.7°C. The final temperature comes to 23.5°C. Calculate the heat capacity of Al in J/mol·°C.

Answer 1

c(calorimeter) = 15.6 J/°C

m(Al) = 24.1 g

t₁(Al) = 100 °C

t₂(Al) = 23.5 °C

m(H₂O) = 99.6 g

c(H₂O) = 4.2 J/g * °C

t₁(H₂O) = 19.7 °C

t₂(H₂O) = 23.5 °C

c(Al) - ?

Solution:

Q(Al) = c(Al) * n(Al) * (t₁ - t₂)

n(Al) = m(Al) / M(Al) = 24.1 / 27 = 0.893 mole

Q(calorimeter with water) = Q(calorimeter) + Q(water)

Q(calorimeter) = c(calorimeter) * (t₂ - t₁) = 15.6 * (23.5 - 19.7) = 59.3 J

Q(water) = c(H₂O) * m(H₂O) * (t₂ - t₁) = 4.2 * 99.6 * (23.5 - 19.7) = 1589.6 J

Q(calorimeter with water) = Q(Al) = c(Al) * n(Al) * (t₁ - t₂)

c(Al) * 0.893 * (100 - 23.5) = 1589.6 + 59.3

c(Al) = 24.1 J / mol * °C

Answer:

c(Al) = 24.1 J / mol * °C