What is the final temperature and physical state of water when 250 grams of water at 85 degrees C is added to 80.0 grams of ice at -15 degree C?

Is the physical state of water liquid?

I still don't understand how to do this.

That doesn't help me. I don't remember if I answered this or not and if I did what I said. If I answered, copy that and explain what you don't understand.

To find the final temperature and physical state of the water, we need to apply the principle of conservation of energy.

First, we need to calculate the heat gained or lost by each substance involved in the process.

1. Heat gained or lost by the water: We can use the equation Q = mcΔT, where Q is the heat gained or lost, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature. The specific heat capacity of water is approximately 4.18 J/g°C.

Q_water = m_water * c_water * ΔT_water
Q_water = 250 g * 4.18 J/g°C * (T_final - 85°C)

2. Heat gained or lost by the ice: The specific heat capacity of ice is negligible compared to water, so we only need to consider the heat of fusion, which is the energy required to convert solid ice into liquid water. The heat of fusion for water is approximately 334 J/g.

Q_ice = m_ice * ΔH_fusion
Q_ice = 80.0 g * 334 J/g

Next, we apply the principle of conservation of energy, which states that the total heat gained is equal to the total heat lost.

Q_water = -Q_ice

Now we can substitute the values into the equation:

250 g * 4.18 J/g°C * (T_final - 85°C) = -80.0 g * 334 J/g

Solving this equation will give us the final temperature of the system.

Once we have the final temperature, we can determine the physical state of water based on the temperature. At temperatures below 0°C, water is in the solid state, and at temperatures above 0°C, water is in the liquid state.