To solve this problem, we can use the equation:
Q_aluminum = -Q_water
where Q_aluminum is the heat lost by the aluminum cube and Q_water is the heat gained by the water. We can also use the specific heat capacity of aluminum and water to calculate the heat lost or gained:
Q = m x c x ΔT
where Q is the heat, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.
First, let's calculate the heat lost by the aluminum cube:
Q_aluminum = m_aluminum x c_aluminum x ΔT_aluminum
Q_aluminum = 1.95 kg x 0.91 J/g°C x (150°C - 60°C)
Q_aluminum = 311.22 kJ
Next, let's calculate the heat gained by the water:
Q_water = m_water x c_water x ΔT_water
Q_water = m_water x 4.18 J/g°C x (27.0°C - 60.0°C)
Q_water = -m_water x 4.18 J/g°C x 33.0°C
Q_water = -1381.14 m_water J
Since Q_aluminum = -Q_water, we can set these equations equal to each other and solve for the mass of water:
311.22 kJ = -1381.14 m_water J
m_water = -311.22 kJ / (-1381.14 J/g°C x 33.0°C)
m_water = 0.666 kg
Therefore, a mass of 0.666 kg of water at 27.0°C must be allowed to come to thermal equilibrium with the aluminum cube to lower its temperature to 60.0°C.