Question

Use the following calorimetric values to answer the question:

The specific heat capacity of water is 4186 J/kg°C.
The specific heat capacity for aluminum is 899 J/kg°C.


A 140-g aluminum ball at 80. °C was dropped into a beaker of water at 10. °C. The final temperature of the water and the ball was 35 °C. What was the mass of the water?
120 g
210 g
290 g
320 g

Answer 1

The mass of the water is approximately 210 g.

Step-by-step explanation:

To find the mass of the water, we can use the principle of conservation of energy. The heat gained by the water and the aluminum ball is equal to the heat lost by the aluminum ball. We can calculate the heat gained by the water using the formula:

Qw = mwcwΔT

Where Qw is the heat gained by the water, mw is the mass of the water, cw is the specific heat capacity of water, and ΔT is the change in temperature. Plugging in the given values, we have:

Qw = (mw)(4186 J/kg°C)(35 °C - 10 °C)

Since we know the mass of the aluminum ball (140 g), we can rearrange the equation to solve for mw:

mw = Qw / (cwΔT)

Substituting the known values, we get:

mw = ((140 g)(899 J/kg°C)(80 °C - 35 °C)) / (4186 J/kg°C)(35 °C - 10 °C)

Simplifying the equation gives us:

mw ≈ 209.55 g ≈ 210 g

Therefore, the mass of the water is approximately 210 g.