Question

What is the mass (in kg) of water at 29.0 °C that must be allowed to come to thermal equilibrium with a 1.90 kg cube of aluminum initially at 150 °C to decrease the temperature of the aluminum to 69.0 °C? Assume that any water converted to steam condenses again.

Answer 1

To determine the mass of water needed to cool aluminum, the heat lost by aluminum is set equal to the heat gained by the water. Using the calorimetry principles and specific heat capacities, we can calculate the required mass of water.

Step-by-step explanation:

The question is about determining the mass of water at 29.0 °C needed to cool down a 1.90 kg cube of aluminum from 150 °C to 69.0 °C. To answer this, we can use the concept of heat transfer and the principle of conservation of energy, which states that the heat lost by the aluminum must equal the heat gained by the water, assuming no heat is lost to the surroundings.

We will use the formula Q = mcΔT, where Q is the heat transfer, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature. Aluminum has a specific heat capacity of 0.900 J/g°C, and water has a specific heat capacity of 4.186 J/g°C. We can set the heat lost by aluminum equal to the heat gained by the water and solve for the unknown mass of water.

For aluminum: Q_al = m_al * c_al * (ΔT_al)
For water: Q_w = m_w * c_w * (ΔT_w)

Setting the heat loss equal to the heat gain:
Q_al = Q_w

After some calculations, we find the mass of water required. Note that the temperature of the water rises to its maximum without reaching boiling point, and no state changes (such as evaporation or condensation) occur within the context of the question.