Question

A student drops two metal objects into a 120 g steel container containing 150 g of water at 25 ° C . One is a 174 g copper cube that is initially at 80 ° C, and the other is a piece of aluminum that is initially at 5.0 ° C . To the student's surprise, the water reaches a final temperature of 25 ° C, precisely as it started . What is the mass (in kg) of the aluminum block ?

Answer 1

We can see here that the mass of the aluminum block is 0 kg.

Let's calculate the heat gained or lost by the copper cube. We can use the equation:

Q = mcΔT

Where Q is the heat transfer, m is the mass of the copper cube, c is the specific heat capacity of copper, and ΔT is the change in temperature.

Given:

• Mass of copper cube (m) = 174 g = 0.174 kg

• Specific heat capacity of copper (c) = 0.386 J/g°C

• Initial temperature of copper (ΔTcopper) = 80 °C

• Final temperature of water (ΔTw) = 25 °C

Using the equation Q = mcΔT, we can calculate the heat gained by the water and lost by the copper cube:

0 = (0.174 kg)(0.386 J/g°C)(25 °C - 80 °C)

Simplifying the equation, we find:

0 = -26.96 J

Since the heat transfer is zero, there is no heat gained or lost by the water or the copper cube. This means that the heat gained by the aluminum block is equal to the heat lost by the copper cube:

0 = ma(0.903 J/g°C)(25 °C - 5.0 °C)

Simplifying the equation, we find:

0 = 20.18ma

Since the heat transfer is zero, we can conclude that the mass of the aluminum block (ma) is also zero.

Therefore, the mass of the aluminum block is 0 kg.