Answer:
Step-by-step explanation:
To solve this problem, we can use the principle of conservation of energy, which states that the total energy of a system is conserved. In this case, the energy transferred between the two samples of water will result in a final temperature that is somewhere between 280 K and 330 K.
We can use the formula:
Q = mcΔT
where Q is the amount of heat transferred, m is the mass of the substance, c is its specific heat capacity, and ΔT is the change in temperature.
Let's assume that the final temperature of the mixture is T. The amount of heat transferred from the hotter water to the cooler water can be calculated as:
Q1 = mcΔT1 = 50.0 g x 4.18 J/(g·K) x (T - 330 K)
where we have used the specific heat capacity of water, which is 4.18 J/(g·K).
Similarly, the amount of heat transferred from the cooler water to the hotter water can be calculated as:
Q2 = mcΔT2 = 30.0 g x 4.18 J/(g·K) x (T - 280 K)
Since the total amount of heat transferred is zero (there is no loss of heat to the surroundings), we can set Q1 equal to -Q2:
50.0 g x 4.18 J/(g·K) x (T - 330 K) = -30.0 g x 4.18 J/(g·K) x (T - 280 K)
Solving for T, we get:
T = (50.0 g x 4.18 J/(g·K) x 330 K + 30.0 g x 4.18 J/(g·K) x 280 K) / (50.0 g x 4.18 J/(g·K) + 30.0 g x 4.18 J/(g·K))
T = 303.9 K
Therefore, the final temperature of the mixture is approximately 303.9 K (or 30.7 °C).