Four ice cubes at exactly 0 ^\circ {\rm{C}} having a total mass of 55.0g are combined with 110g of water at 77^\circ C in an insulated container.
Part A
If no heat is lost to the surroundings, what will be the final temperature of the mixture?
I thought I did this earlier.
To find the final temperature of the mixture, we need to use the principle of conservation of energy. The heat lost by the hot water will be equal to the heat gained by the ice cubes.
The equation for heat transfer is given by:
Q = m * c * ΔT
Where:
Q is the heat transfer (in Joules),
m is the mass of the substance (in grams),
c is the specific heat capacity of the substance (in J/g°C), and
ΔT is the change in temperature (in °C).
First, let's calculate the heat lost by the hot water.
The mass of the water is 110g, and the initial temperature is 77°C. We take the final temperature as the unknown variable. So the change in temperature for the water is:
ΔT_water = final temperature - initial temperature
Next, let's calculate the heat gained by the ice cubes.
The mass of the ice cubes is 55.0g, and their initial temperature is 0°C. We take the final temperature as the unknown variable. So the change in temperature for the ice cubes is:
ΔT_ice = final temperature - initial temperature
Since no heat is lost to the surroundings, the heat lost by the hot water is equal to the heat gained by the ice cubes. Therefore, we can set up the equation:
m_water * c_water * ΔT_water = m_ice * c_ice * ΔT_ice
Substituting the known values:
110g * c_water * ΔT_water = 55.0g * c_ice * ΔT_ice
Now, we need to find the specific heat capacities (c) for water and ice. The specific heat capacity of water is approximately 4.18 J/g°C, and for ice, it is approximately 2.09 J/g°C.
Substituting these values and rearranging the equation:
110g * 4.18 J/g°C * ΔT_water = 55.0g * 2.09 J/g°C * ΔT_ice
Simplifying the equation:
ΔT_water = (55.0g / 110g) * (2.09 J/g°C / 4.18 J/g°C) * ΔT_ice
ΔT_water = 0.5 * 0.5 * ΔT_ice
ΔT_water = 0.25 * ΔT_ice
Since ΔT_water + ΔT_ice = 77°C (the initial temperature difference between the water and ice), we can substitute this value into the equation:
0.25 * ΔT_ice + ΔT_ice = 77°C
1.25 * ΔT_ice = 77°C
ΔT_ice ≈ 61.6°C
Finally, we can substitute the value of ΔT_ice into ΔT_water to find the final temperature of the mixture:
ΔT_water = 0.25 * 61.6°C
ΔT_water ≈ 15.4°C
So, the final temperature of the mixture will be:
Final temperature = initial temperature of water + ΔT_water
Final temperature = 77°C + 15.4°C
Final temperature ≈ 92.4°C