what is the final temp of the combination when 50 grams of chromium at 15 degrees celcius is added to 25 milliliters of water at 45 degrees celcius?
Note the correct spelling of celsius.
[mass Cr x specific heat Cr x (Tfinal-Tinitial)] + [mass water x spcific heat water x (Tfinal-Tinitial) = 0
Solve for Tfinal.
Post your work if you get stuck.
(50g)(0.107 cal/g degreesC)(tfinal-15C) +(25g)(1.000 cal/g degreesC)(tfinal-45C) is this correct
To find the final temperature of the combination, we can use the principle of conservation of energy. The formula we will be using is:
(q1 + q2) = mcΔT
Where:
q1 = heat gained or lost by the first substance
q2 = heat gained or lost by the second substance
m = mass of the substance
c = specific heat capacity of the substance
ΔT = change in temperature
First, let's find the heat gained or lost by chromium (q1). We can use the formula:
q1 = mcΔT
The mass of chromium (m1) is 50 grams, the specific heat capacity of chromium (c1) is 0.71 J/g°C, and the change in temperature (ΔT1) is the final temperature minus the initial temperature. Since we're trying to find the final temperature, we can represent it as T:
q1 = (50 g) x (0.71 J/g°C) x (T - 15°C)
Next, let's find the heat gained or lost by water (q2). We can use the formula:
q2 = mcΔT
The mass of water (m2) is 25 grams (since 1 milliliter of water is approximately 1 gram), the specific heat capacity of water (c2) is 4.18 J/g°C, and the change in temperature (ΔT2) is the final temperature minus the initial temperature. In this case, the initial temperature is 45°C:
q2 = (25 g) x (4.18 J/g°C) x (T - 45°C)
Now, we can equate q1 and q2, and solve for T:
(50 g) x (0.71 J/g°C) x (T - 15°C) = (25 g) x (4.18 J/g°C) x (T - 45°C)
Simplifying:
35(T - 15) = 104.5(T - 45)
35T - 525 = 104.5T - 2347.5
69.5T = 1822.5
T ≈ 26.22°C
Therefore, the final temperature of the combination of chromium and water is approximately 26.22°C.