To solve this problem, we can use the principle of conservation of energy and assume that there is no heat loss to the surroundings.
First, let's calculate the heat gained by the ice when it reaches thermal equilibrium with the water. We can use the equation:
Q_ice = m_ice * c_ice * ΔT_ice,
where Q_ice is the heat gained by the ice, m_ice is the mass of the ice, c_ice is the specific heat capacity of ice, and ΔT_ice is the change in temperature of the ice.
The specific heat capacity of ice is approximately 2.09 J/(g·°C).
Given: m_ice = 60 kg = 60000 g ΔT_ice = 0°C - (-10°C) = 10°C
Q_ice = 60000 g * 2.09 J/(g·°C) * 10°C = 1254000 J
Now, let's calculate the heat lost by the water when it reaches thermal equilibrium with the ice:
Q_water = m_water * c_water * ΔT_water,
where Q_water is the heat lost by the water, m_water is the mass of the water, c_water is the specific heat capacity of water, and ΔT_water is the change in temperature of the water.
The specific heat capacity of water is approximately 4.18 J/(g·°C).
Given: m_water = 0.30 l = 300 g (since 1 g of water is approximately equal to 1 ml) ΔT_water = 50°C - 0°C = 50°C
Q_water = 300 g * 4.18 J/(g·°C) * 50°C = 62700 J
Since there is no heat loss to the surroundings, the heat gained by the ice is equal to the heat lost by the water:
Q_ice = Q_water
1254000 J = 62700 J + m_final * c_water * ΔT_final,
where m_final is the mass of the final mixture and ΔT_final is the change in temperature of the final mixture.
We know that the initial temperature of the ice is -10°C, and the final temperature of the mixture is the same since it has reached thermal equilibrium.
Therefore, ΔT_final = 0°C - (-10°C) = 10°C.
Simplifying the equation, we can solve for m_final:
1254000 J = 62700 J + m_final * 4.18 J/(g·°C) * 10°C
1191300 J = m_final * 41.8 J/g
m_final = 1191300 J / 41.8 J/g ≈ 28530 g ≈ 28.53 kg
Therefore, when the system reaches thermal equilibrium, the total mass of the mixture (ice + water) is approximately 28.53 kg.