A piece of gold with a mass of 21.0 g was heated to 100.0 °C and placed inside a coffee cup calorimeter. The temperature of the water inside the calorimeter increased from 23.8 °C to 25.5 °C. The specific heat capacity of water is 4.184 J g^-1 °C^-1 and the specific heat capacity of gold is 0.129 J g^-1 °C^-1. Calculate the mass of water contained inside the calorimeter.

Question

Quizplus · Accepted Answer

First, we need to use the equation q = mcΔT to find the amount of heat transferred from the gold to the water in the calorimeter. Since no heat is lost to the environment, we can assume that the heat lost by the gold is equal to the heat gained by the water:

q(gold) = -q(water)

m(gold) x c(gold) x ΔT(gold) = -m(water) x c(water) x ΔT(water)

(21.0 g) x (0.129 J/g°C) x (100.0 - 25.5) °C = -(m(water)) x (4.184 J/g°C) x (25.5 - 23.8) °C

-226.2 J = -(m(water)) x 35.8 J

m(water) = 6.32 g

However, this is only the mass of water that was used to raise its own temperature. We need to account for the total mass of water in the calorimeter, which includes the initial mass of water and the mass of the hot gold that was added. Let's call the mass of water we just calculated "m1":

Total mass of water = m1 + m2

We can use the fact that the initial temperature of the water was 23.8 °C to find the mass of water that was initially in the calorimeter:

q(water) = m(water) x c(water) x ΔT(water)

q(water) = (m1 + m2) x (4.184 J/g°C) x (25.5 - 23.8) °C

q(water) = 226.2 J

Solving for m2, the mass of the water initially in the calorimeter:
226.2 J = m2 x 4.184 J/g°C x (25.5 - 23.8) °C
m2 = 28.4 g

Therefore, the total mass of water in the calorimeter is:
m1 + m2 = 6.32 g + 28.4 g = 34.7 g, which is closest to answer choice A.

A Piece of Gold with a Mass of 21