Question

A sample of quartz is put into a calorimeter (see sketch at right) that contains of water. The quartz sample starts off at and the temperature of the water starts off at . When the temperature of the water stops changing it's . The pressure remains constant at . Calculate the specific heat capacity of quartz according to this experiment. Be sure your answer is rounded to the correct number of significant digits.

Answer 1

Answer:

0.71 J/g°C

Explanation:

Here is the complete question

thermometer A 51.9 g sample of quartz is put into a calorimeter (see sketch at right) that contains 300.0 g of water. The quartz sample starts off at 97.8 °C and the temperature of the water starts off at 17.0 °C. When the temperature of the water stops changing it's 19.3 °C. The pressure remains constant at 1 atm. insulated container water sample Calculate the specific heat capacity of quartz according to this experiment. Be sure your answer is rounded to 2 significant digits. a calorimeter g °C

Solution

Since the temperature of the water increases from 17.0 °C to 19.3 °C, it means that it loses heat. Also, the final temperature of the quartz equals the final temperature of the water 19.3 °C. Since the quartz temperature decreases from 97.8 °C to 19.3 °C it loses heat.

So, heat lost by quartz, Q = heat gained by water, Q'

-Q = Q'

-mc(θ₂ - θ₁) = m'c'(θ₂ - θ₃) where m = mass of quartz = 51.9 g, c = specific heat capacity of quartz, θ₁ = initial temperature of quartz = 97.8 °C, θ₂ = final temperature of quartz = 19.3 °C, m' = mass of water = 300 g, c = specific heat capacity of water = 4.2 J/g °C , θ₃ = initial temperature of water = 17.0 °C, θ₂ = final temperature of water = 19.3 °C

Making c subject of the formula, we have

c = -m'c'(θ₂ - θ₃)/m(θ₂ - θ₁)

Substituting the values of the variables into the equation, we have

c = -300 g × 4.2 J/g °C(19.3 °C - 17.0 °C)/51.9 g(19.3 °C - 97.8 °C)

c = -1260 J/°C(2.3 °C)/51.9 g(-78.5 °C)

c = -2898 J/-4074.15 g°C

c = 0.711 J/g°C

c ≅ 0.71 J/g°C to 2 significant digits