A 54.7-g sample of aluminum at 95.0°C is dropped into 35.0 g of water at 40.0°C. What is the final temperature of the mixture? (specific heat capacity of aluminum = 0.89 J/g°C; specific heat capacity of water = 4.184 J/g°C)
Summation of all heat is zero.
Q,aluminum + Q,water = 0
The formula for Q is
Q = mc(T2-T1)
where
m = mass (g)
c = specific heat capacity (J/g-K)
T = temperature (K)
Substitute the values. Note that the final Temperature (T2) is the same for both:
[mc(T2-T1)],aluminum + [mc(T2-T1)],water = 0
54.7 * 0.89 * (T2 - 95) + 35 * 4.184 * (T2 - 40) = 0
Now solve for T2. The value you should get must be between 40 and 95 °C.
Hope this helps~ `u`
Well, we have aluminum getting cozy with water, quite the party going on! Let's see if we can calculate the final temperature and join in on the fun.
To start, we need to calculate the heat gained or lost by each substance.
For the aluminum, we use the formula Q = m * c * ΔT, 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.
So for the aluminum, we have:
Q_aluminum = (54.7 g) * (0.89 J/g°C) * (T_final - 95.0°C)
For the water, we have:
Q_water = (35.0 g) * (4.184 J/g°C) * (T_final - 40.0°C)
Now, according to the law of conservation of energy, the heat lost by the aluminum equals the heat gained by the water, which means:
Q_aluminum = -Q_water
Let's plug our values in and solve the equation:
(54.7 g) * (0.89 J/g°C) * (T_final - 95.0°C) = -(35.0 g) * (4.184 J/g°C) * (T_final - 40.0°C)
Now, we can simplify and solve for T_final:
(48.683 J/°C) * (T_final - 95.0°C) = -(146.44 J/°C) * (T_final - 40.0°C)
Okay, let's simplify further:
48.683 T_final - 4605.57 = -146.44 T_final + 5857.6
Add 146.44 T_final to both sides and subtract 4605.57 from both sides:
195.123 T_final = 10463.17
Divide both sides by 195.123:
T_final ≈ 53.61°C
Voila! The final temperature of the mixture should be around 53.61°C. Time to jump in and join the aluminum and water's pool party! But be careful, aluminum and water don't mix well in reality, they might prefer to stay separate.
To solve this problem, we can use the principle of conservation of energy.
Step 1: Calculate the heat absorbed or released by the aluminum.
To do this, we can use the formula:
q(aluminum) = mass x specific heat capacity x change in temperature
q(aluminum) = 54.7 g x 0.89 J/g°C x (final temperature - 95.0°C)
Step 2: Calculate the heat absorbed or released by the water.
Using the same formula:
q(water) = mass x specific heat capacity x change in temperature
q(water) = 35.0 g x 4.184 J/g°C x (final temperature - 40.0°C)
Step 3: Apply the principle of conservation of energy.
According to the principle, the heat absorbed by the aluminum is equal to the heat released by the water. Therefore:
q(aluminum) = - q(water)
54.7 g x 0.89 J/g°C x (final temperature - 95.0°C) = -35.0 g x 4.184 J/g°C x (final temperature - 40.0°C)
Step 4: Simplify and solve for the final temperature.
Solving for the final temperature:
48.633 x (final temperature - 95.0) = -146.24 x (final temperature - 40.0)
48.633 x final temperature - 48.633 x 95.0 = -146.24 x final temperature + 146.24 x 40.0
48.633 x final temperature + 146.24 x final temperature = 146.24 x 40.0 + 48.633 x 95.0
194.873 x final temperature = 5849.6 + 4618.335
194.873 x final temperature = 10467.935
final temperature = 10467.935/194.873
final temperature ≈ 53.7°C
Therefore, the final temperature of the mixture is approximately 53.7°C.
To find the final temperature of the mixture, we can use the principle of conservation of energy.
The heat gained by the water will be equal to the heat lost by the aluminum. The formula to calculate the heat gained or lost is:
q = m * c * ΔT
Where:
q = heat gained or lost
m = mass of the substance
c = specific heat capacity of the substance
ΔT = change in temperature
For the water:
Mass (m1) = 35.0 g
Specific heat capacity (c1) = 4.184 J/g°C
Initial temperature (T1) = 40.0°C
Final temperature (T2) = ?
For the aluminum:
Mass (m2) = 54.7 g
Specific heat capacity (c2) = 0.89 J/g°C
Initial temperature (T3) = 95.0°C
Final temperature (T2) = ?
Using the formula for the heat gained or lost, we have:
Heat gained by the water (q1) = m1 * c1 * (T2 - T1)
Heat lost by the aluminum (q2) = m2 * c2 * (T3 - T2)
Since the total heat gained by the water is equal to the total heat lost by the aluminum, we can set up the equation:
q1 = q2
m1 * c1 * (T2 - T1) = m2 * c2 * (T3 - T2)
Now we can substitute the given values into the equation and solve for T2, the final temperature of the mixture:
35.0 g * 4.184 J/g°C * (T2 - 40.0°C) = 54.7 g * 0.89 J/g°C * (95.0°C - T2)
Simplifying the equation:
146.84(T2 - 40.0) = 48.683(95.0 - T2)
Solving for T2:
146.84T2 - 146.84(40.0) = 48.683(95.0) - 48.683T2
146.84T2 + 48.683T2 = 48.683(95.0) + 146.84(40.0)
195.523T2 = 4634.535 + 5873.6
195.523T2 = 10508.135
T2 = 10508.135 / 195.523
T2 ≈ 53.7°C
Therefore, the final temperature of the mixture is approximately 53.7°C.