IF 100.0G OF WATER AT 90'C IS ADDED TO 50.0G OF WATER AT 10'C, ESTIMATE THE FIBNAL TEMPATURE OF WATER. EXPLAIN YOUR REASONING.
PLEASE HELP I DON'T GET IT AND I HAVE ANOTHER PROBLEM SIMILAR TO THIS
heat lost by warm water + heat gained by cold water = 0\
[mass warm water x specific heat H2O x (Tfinal-Tinitial)] + [mass cool water x specific heat H2O x (Tfinal-Tinitial)] = 0
Substitute and solve for Tf.
Post your work if you get stuck.
To estimate the final temperature of the water after mixing, we can use the principle of conservation of energy. When the two bodies of water come into contact, heat energy will flow from the hotter water to the cooler water until both reach thermal equilibrium.
To solve this problem, we need to calculate the heat gained and lost by each body of water, and then set them equal to each other.
First, let's calculate the heat gained (or lost) by the water at 90°C using the formula:
Q = m * c * ΔT
Where:
Q = Heat gained or lost
m = Mass of the water
c = Specific heat capacity of water (4.18 J/g°C)
ΔT = Change in temperature
The heat gained by the water at 90°C can be calculated as:
Q1 = (100.0 g) * (4.18 J/g°C) * (Tf - 90°C)
Next, let's calculate the heat gained (or lost) by the water at 10°C:
Q2 = (50.0 g) * (4.18 J/g°C) * (Tf - 10°C)
According to the principle of conservation of energy, the heat gained by the colder water must be equal to the heat lost by the hotter water. Therefore, we can set Q1 equal to Q2:
Q1 = Q2
(100.0 g) * (4.18 J/g°C) * (Tf - 90°C) = (50.0 g) * (4.18 J/g°C) * (Tf - 10°C)
Now, you can solve the above equation to find the final temperature (Tf) of the water.
To estimate the final temperature of the water, we can use the principle of conservation of energy, specifically the law of heat exchange. The equation we will use is:
Heat gained by cold water = Heat lost by hot water
The amount of heat gained or lost by an object is given by the equation:
Q = mcΔT
Where:
Q is the heat gained or lost
m is the mass
c is the specific heat capacity
ΔT is the change in temperature
In this case, we have two masses of water with different temperatures. Let's calculate the heat lost by the hot water first:
Heat lost by hot water = m1 * c1 * ΔT1
Given:
m1 = 100.0 g (mass of hot water)
c1 = 4.18 J/g°C (specific heat capacity of water)
ΔT1 = (final temperature - 90°C)
Next, let's calculate the heat gained by the cold water:
Heat gained by cold water = m2 * c2 * ΔT2
Given:
m2 = 50.0 g (mass of cold water)
c2 = 4.18 J/g°C (specific heat capacity of water)
ΔT2 = (final temperature - 10°C)
Now, since heat lost by the hot water is equal to heat gained by the cold water, we can set up the equation:
m1 * c1 * ΔT1 = m2 * c2 * ΔT2
Plugging in the given values, we have:
100.0 * 4.18 * (final temperature - 90) = 50.0 * 4.18 * (final temperature - 10)
Now, we can solve this equation to find the final temperature:
418 * (final temperature - 90) = 209 * (final temperature - 10)
Simplifying:
418 * final temperature - 37620 = 209 * final temperature - 2090
209 * final temperature - 418 * final temperature = 2090 - 37620
(-209 * final temperature) = -35530
Dividing both sides by -209:
final temperature = 170°C
So, the estimated final temperature of the water is 170°C.