What would be the final temperature of a mixture of 50 g{\rm g} of 17∘C^\circ C water and 50 g{\rm g} of 44∘C^\circ C water?
I have no clue where I should even begin please help.
The sum of the heats gained is zero.
HeatONE+HeatTWO=0
50*c*(Tf-17)+50*c*(Tf-44)=0
c divides out, solve for Tf. Actually, 50 divides out, so Tf= (17+44)/2
To find the final temperature of a mixture of two substances, we can use the principle of conservation of energy, specifically heat transfer. The total heat gained by one substance is equal to the total heat lost by the other substance.
To solve this problem, we can use the formula:
q1 + q2 = 0
where q1 is the heat gained by the first substance (17°C water) and q2 is the heat gained by the second substance (44°C water).
The equation for heat transfer is:
q = mcΔT
where q represents the heat transferred, m is the mass of the substance, c is the specific heat capacity of the substance, and ΔT is the change in temperature.
First, let's calculate the heat gained by the 17°C water (q1). We know the mass of the water is 50 g and the initial temperature is 17°C. The specific heat capacity of water is approximately 4.18 J/g°C.
q1 = mcΔT
= (50 g)(4.18 J/g°C)(Tf - 17°C)
Next, let's calculate the heat gained by the 44°C water (q2). The mass of the water is 50 g and the initial temperature is 44°C.
q2 = mcΔT
= (50 g)(4.18 J/g°C)(44°C - Tf)
Since the total heat gained and lost must balance, we can set up the equation:
q1 + q2 = 0
(50 g)(4.18 J/g°C)(Tf - 17°C) + (50 g)(4.18 J/g°C)(44°C - Tf) = 0
Now we can solve for Tf, the final temperature of the mixture.