95.0 g of0.0 degrees celsius ice is added to exactly 100 g of water at 60.0 degrees celsius. when the temperature of the mixture first reaches 0.0 celsius the mass of the ice present is?
To find the mass of the ice present when the temperature of the mixture reaches 0.0 degrees Celsius, we need to use the principle of conservation of energy.
When the ice and water are mixed, heat energy is transferred from the water to the ice in order to melt the ice and raise its temperature to 0.0 degrees Celsius. The equation we can use is:
Q_water + Q_ice = 0
Where Q_water is the heat gained by the water and Q_ice is the heat lost by the ice.
The formula to calculate the heat gained or lost by a substance is given by:
Q = m * c * ΔT
Where Q is the heat gained or lost, m is the mass of the substance, c is the specific heat capacity of the substance, and ΔT is the change in temperature.
For water, the specific heat capacity (c) is approximately 4.18 J/g°C, and for ice, it is approximately 2.09 J/g°C.
Let's calculate the heat gained by the water:
Q_water = m_water * c_water * ΔT_water
= 100 g * 4.18 J/g°C * (0.0°C - 60.0°C) [since the water is at 60.0 degrees Celsius initially]
Q_water = -25080 J
Now, let's calculate the heat lost by the ice:
Q_ice = m_ice * c_ice * ΔT_ice
= m_ice * 2.09 J/g°C * (0.0°C - 0.0°C) [since the ice is initially at 0.0 degrees Celsius]
Since the ice is at its melting point, it will change phase (from solid to liquid) instead of changing temperature. This phase change requires energy, which is known as the heat of fusion.
The heat of fusion for water is approximately 334 J/g. So, the heat lost by the ice during melting can be calculated as:
Q_ice = m_ice * 334 J/g
Now, since the total heat gained by the water must be equal to the total heat lost by the ice:
Q_water + Q_ice = 0
-25080 J + m_ice * 334 J/g = 0
Solving for m_ice (mass of ice):
m_ice = 25080 J / 334 J/g
≈ 75.15 g
Therefore, when the temperature of the mixture first reaches 0.0 degrees Celsius, the mass of the ice present is approximately 75.15 grams.
Temperature of water must decrease by 60C; how much energy is lost by the water in doing so?
mass H2O x specific heat H2O x 60 = approximately 25,000
How much can we get from 95g ice melting? That is
95 x 334 J/g =about 32,000
The differeence is 32,000-25,000 = about 7,000 and 7000/334 = 21. Approximately 21 g ice did not melt.
I estimated here and there; you need to go through and redo the problem more accurately.