If 118 g of ice at 0.0°C is added to 1.55 L of water at 88°C, what is the final temperature of the mixture?
heat absorbed by melting ice + heat to raise water from zero C + heat lost by 88 C water.= 0
[mass ice x heat fusion] + [mass melted ice x specific heat water x (Tfinal-Tinitial)] + [mass separate water x specific heat water x (Tfinal-Tinitial)] = 0.
Substitute the numbers and solve for Tfinal.
how do i get the mass of separate water, if it is in L?
1.55L is 1550 mL (1550 cc) and if the density of water is 1.00 g/mL, then the mass of the water is 1550 grams. Most problems of this nature usually state that the water is to be considered as having a density of 1 g/cc.
um...one more question, what do you mean by "mass of melted ice"? how is it determined?
Let's see now. If I had 1 g ice and it melted into water, it would form 1 g water. So grams ice and grams melted ice must be the same. right?
To determine the final temperature of the mixture, we can use the principle of energy conservation. The heat lost by the hot water will be equal to the heat gained by the ice to reach the final temperature.
First, let's calculate the heat lost by the hot water using the formula:
Q = m * c * ΔT
Where:
Q is the heat lost
m is the mass of the water
c is the specific heat capacity of water
ΔT is the change in temperature
Given:
Mass of water (m) = 1.55 L
Density of water = 1 g/cm³
1 L of water = 1 kg
1 kg of water = 1000 g
So, mass of water (m) = 1550 g
Specific heat capacity of water (c) = 4.186 J/g°C
Change in temperature (ΔT) = Final temperature - Initial temperature
The initial temperature of the water is 88°C.
Now, let's calculate the heat gained by the ice. To melt the ice, the heat absorbed can be calculated using the formula:
Q = m * ΔHf
Where:
Q is the heat gained
m is the mass of the ice
ΔHf is the heat of fusion of ice
Given:
Mass of ice = 118 g
Heat of fusion of water (ΔHf) = 334 J/g
Now, let's equate the heat lost by the hot water to the heat gained by melting the ice:
Q (water) = Q (ice)
m (water) * c (water) * ΔT = m (ice) * ΔHf
Substituting the known values:
1550 g * 4.186 J/g°C * (Final temperature - 88°C) = 118 g * 334 J/g
Simplifying the equation:
1550 * 4.186 * (Final temperature - 88) = 118 * 334
Now we can solve for the final temperature:
((1550 * 4.186 * Final temperature) - (1550 * 4.186 * 88)) = (118 * 334)
6212.87 * Final temperature - (623192) = 39392
6212.87 * Final temperature = 662584
Final temperature = 106.64°C
Therefore, the final temperature of the mixture is approximately 106.64°C.