if a 100g sample of water at 60.0 C is added to a 100.0 g sample of water at 10.0 C, determine the final temperature of the water. Assume no heat is lost to the surroundings
The equation to be used for the problem is mcΔT = mcΔT.
From the given problem, we need to determine whether the heat is absorbed or evolved. When heat is absorbed, the sign is positive, and when heat is evolved, it is negative; therefore +q = -q, due to the reason then when the 100g of sample at 60.0C was added, it absorbed heat while the 100.0g sample of water at 10.0C evolved the heat towards the previous sample.
Applying the equation with the proper signs, we get mcΔT = -mcΔT, the specific heat of water is cancelled out right away.
Required: Final Temperature (T2)
mcΔT = (1)
-mcΔT = (2)
(1) m = 100.00g
T2-60.0C
(2) m = 100.00g
T2-10.0C
(100g) (T2-60.0C) = -(100g) (T2-10.0C)
Multiply the mass with the (T2-T1) equation.
100gT2 - 6000C = -100gT2 - 1000C
Transpose the values.
100gT2+100gT2 = 6000C - 1000C
200gT2 = 5000C
T2 = 5000C/200g
T2 = 25C.
62.5
To determine the final temperature of the water, we can use the principle of heat transfer. The heat lost by the warmer water will be equal to the heat gained by the cooler water.
To calculate the heat gained or lost, we can use the formula:
Q = m * C * ΔT
Where:
Q is the heat gained or lost
m is the mass of the water
C is the specific heat capacity of water (4.18 J/g°C)
ΔT is the change in temperature
First, let's calculate the heat lost by the warmer water (Q1).
ΔT1 = T1 - Tf
(where T1 is the initial temperature of the warmer water and Tf is the final temperature of the water)
Q1 = m1 * C * ΔT1
m1 = mass of the warmer water = 100 g
T1 = initial temperature of the warmer water = 60.0 °C
Next, let's calculate the heat gained by the cooler water (Q2)
ΔT2 = Tf - T2
(where T2 is the initial temperature of the cooler water and Tf is the final temperature of the water)
Q2 = m2 * C * ΔT2
m2 = mass of the cooler water = 100 g
T2 = initial temperature of the cooler water = 10.0 °C
Since the heat lost by the warmer water is equal to the heat gained by the cooler water, we can equate Q1 and Q2:
m1 * C * ΔT1 = m2 * C * ΔT2
Now, let's plug in the values and solve for Tf, the final temperature:
100 g * 4.18 J/g°C * (60.0 °C - Tf) = 100 g * 4.18 J/g°C * (Tf - 10.0 °C)
Simplifying the equation:
(60.0 °C - Tf) = (Tf - 10.0 °C)
60.0 °C - Tf = Tf - 10.0 °C
Combining like terms:
60.0 °C + 10.0 °C = Tf + Tf
70.0 °C = 2Tf
Dividing by 2:
Tf = 70.0 °C / 2
Tf = 35.0 °C
Therefore, the final temperature of the water will be 35.0 °C.