andy mixed 40 ml of water at 15 degress celius with 40 ml of water at 65 degrees celius ..... what is the final temperature of the water.?? H E L P !
since they are the same masses of water, the average temp is the answer
(40+15)/2 = 55/2 = 27 1/2
Thank you soo much damon !!
To find the final temperature of the water mixture, we can use the principle of heat transfer, known as the heat equation:
Q = m * c * ΔT
Where:
Q is the heat transferred
m is the mass of the substance
c is the specific heat capacity of the substance
ΔT is the change in temperature
In this case, we have two equal masses of water, 40 ml each, so the mass of the water is 80 grams since 1 ml of water is equal to 1 gram. We will assume the specific heat capacity of water, c, to be 4.18 J/g°C.
Let's calculate the heat transferred from the water at 65 degrees Celsius to the water at 15 degrees Celsius:
Q1 = m * c * ΔT
Q2 = m * c * ΔT
Since the masses and the specific heat capacity are the same, we can simplify the equation:
Q1 = Q2
Q1 = Q2 = Q
So, the heat transferred from the water at 65 degrees Celsius to the water at 15 degrees Celsius is equal.
To find the final temperature, we need to equate the two equations:
m * c * ΔT1 = m * c * ΔT2
m * c * ΔT1 - m * c * ΔT2 = 0
80 * (15 - T) - 80 * (T - 65) = 0 [replacing ΔT1 with (15 - T) and ΔT2 with (T - 65)]
1200 - 80T - 80T + 5200 = 0 [expanding the equation]
-160T + 6400 = 0 [combining like terms]
-160T = -6400 [moving constants to the right side]
T = -6400 / -160 [dividing both sides by -160]
T = 40
Therefore, the final temperature of the water mixture will be 40 degrees Celsius.