Dani has 15.00 grams of water (c = 4.186 joules over grams times degrees Celsius) at 66°C. She adds 40.00 grams of water at 35°C.
What is the final temperature of the liquids? Express the answer to the hundredths place.
To find the final temperature of the liquids, we can use the principle of conservation of energy. The total amount of energy (heat) transferred between the two liquids can be calculated using the equation:
Q = m1 * c * ΔT1 + m2 * c * ΔT2
Where:
Q = Heat transferred
m1 = mass of the first liquid (15.00 grams)
c = specific heat capacity (4.186 J/g°C)
ΔT1 = change in temperature of the first liquid (final temperature - initial temperature)
m2 = mass of the second liquid (40.00 grams)
ΔT2 = change in temperature of the second liquid (final temperature - initial temperature)
Since we want to find the final temperature, let's rewrite the equation as:
Q = m1 * c * (Tf - 66) + m2 * c * (Tf - 35)
We know that Q (heat transferred) is zero because no energy is lost or gained by the system. Thus:
0 = m1 * c * (Tf - 66) + m2 * c * (Tf - 35)
Simplifying the equation:
0 = 15.00 * 4.186 * (Tf - 66) + 40.00 * 4.186 * (Tf - 35)
Now, let's solve for Tf:
0 = 62.79(Tf - 66) + 167.44(Tf - 35)
Expanding and rearranging the equation:
0 = 62.79Tf - 4146.54 + 167.44Tf - 5862.80
0 = 230.23Tf - 10009.34
Solving for Tf:
230.23Tf = 10009.34
Tf = 10009.34 / 230.23 ≈ 43.47
Therefore, the final temperature of the liquids is approximately 43.47°C (or you can round to the hundredths place as 43.47).