A 30.0 gram sample of water at 280. K is mixed with 50.0 grams of water at 330. K. Calculate the final temperature of the mixture assuming no heat loss to the surroundings.
heat gained by cold water + heat lost by hot water = 0
mass1 H2O x specific heat water x (Tfinal-Tinitial)] + [mass2 H2O x specific heat water x (Tfinal-Tintial)] = 0
To calculate the final temperature of the mixture, we need to apply the principle of heat transfer known as the law of conservation of energy. According to this principle, the total heat gained by one part of a system should equal the total heat lost by another part of the system.
To solve this problem, we can use the formula:
q1 + q2 = 0
where q1 is the heat gained or lost by the first part (30.0 grams of water at 280 K) and q2 is the heat gained or lost by the second part (50.0 grams of water at 330 K).
First, let's calculate q1:
q1 = m1 * c * ΔT1
where
m1 is the mass of the first part (30.0 grams),
c is the specific heat capacity of water (4.18 J/g·K), and
ΔT1 is the change in temperature of the first part (final temperature - initial temperature).
Next, let's calculate q2:
q2 = m2 * c * ΔT2
where
m2 is the mass of the second part (50.0 grams),
c is the specific heat capacity of water (4.18 J/g·K), and
ΔT2 is the change in temperature of the second part (final temperature - initial temperature).
Since both parts are brought to the same final temperature, the change in temperature is the same for both parts. Let's represent the final temperature as T:
ΔT1 = T - 280
ΔT2 = T - 330
Now, we can substitute these values into the heat transfer equations:
q1 = 30.0 * 4.18 * (T - 280)
q2 = 50.0 * 4.18 * (T - 330)
Since q1 + q2 = 0, we can set up the equation:
30.0 * 4.18 * (T - 280) + 50.0 * 4.18 * (T - 330) = 0
Now, we can solve this equation to find the final temperature T.