Practice Question 1/17
If you mix 200g of Hot water at 80 C with 50g of Cold water at 20 C, what will be the final temperature of the
mixture?
What we know:
Remember: Heat lost by hot water = Heat gained by cold water
And we know that H = m * T
where H = heat in calories, m = mass and T = change in Temperature (or Final T – Starting T)
What we were given:
For hot water: m = 200g
For cold water: m = 50g
Starting Temp = 80C
Starting Temp = 20C
What we what to know = final temperature of the cold & hot water mixture.
There are many ways to calculate this, one is listed below:
Step #1:
Set up a table as Follows
Remember: Hot
water starts with T =
80C
Heat
transferred
T for Hot
Water
New T for Hot
Water
Remember: Cold water
starts with T = 20C
T for Cold
Water
New T for Cold
water
Step #1 pick an amount of heat to transfer from the hot to the cold water. For this 1st step any amount of heat
will work. You can choose 1 calorie, 50 calories, 100 calories, 1000 calories, etc. However, if you are not sure
where to start a good place to begin is to choose an amount of heat that will change the larger water mass by
1C.
Let’s calculate this:
The larger water mass = 200 g, so using
H = m*T
H = 200g * 1C
H = 200gC
or since 1 calorie = 1 gC
H = 200 calories.
Now plug these value into your table:
Remember: Hot water
starts with T = 80C
Heat
transferred
200 calories
T for Hot
Water
1C
New T for Hot
Water
80C - 1C = 79C
Remember: Cold water
starts with T = 20C
T for Cold
Water
New T for Cold
water
Step #2: Calculate how much the cold water changed as a result of gaining that same 200 calories:
H = m*T(cold)
200 cal = 50g*T(cold)
200 cal / 50g = T(cold)
4C
= T(cold)
Now plug this value into your table
Remember: Hot water
starts with T = 80C
Heat
transferred
200 calories
T for Hot
Water
1C
New T for Hot
Water
80C - 1C = 79C
Remember: Cold water
starts with T = 20C
T for Cold
Water
4C
New T for Cold
water
20C + 4C = 24C
If the New T Hot Water = New T Cold Water, then you are done.
If: New T Hot Water ≠ New T Cold Water you need to keep going (keep in mind that if the new value for the
“cold water” will never be greater than the new value for your “hot water”).
In this case, New T Hot Water ≠ New T Cold Water, so we need to keep going
Step #3: Pick a new amount of heat to transfer and complete another line on your table. Again, you can choose
any amount. In this case we can save some time by increasing the amount of heat transferred. Let’s try 2000
calories instead of 2000.
Using the same method as described above, we 1st calculate the new temperature of the hot and cold
water:
For the Hot Water:
Heat = m*T(hot)
2000 cal
= 200g*T(hot)
2000 cal / 200g = T(hot)
10C = T(hot)
For the Cold Water:
Heat = m*T(hot)
2000 cal = 50g*T(cold)
2000 cal / 50g = T(cold)
40C = T(cold)
Use these values to fill in the next line of your table.
Remember: Hot water starts
with T = 80C
Heat
transferred
200 calories
T for Hot
Water
1C
New T for Hot Water
80C - 1C = 79C
Remember: Cold water starts
with T = 20C
T for Cold
Water
4C
Use new Hot Water T from
the line above = 79C
2000 calories
10C
79C - 10C = 69C
New T for Cold water
20C + 4C = 24C
Use new Cot Water T from
the line above = 24C
40C
24C + 40C = 64C
We are now close, but New T Hot Water ≠ New T Cold Water, so we need to keep going. Since we are close,
let’s go back to transferring a smaller amount of heat (200 cal).
Step #4:
Form steps 1 & 2, we know that when 200calories are tranfered, T(hot) = 1C and T(cold) =4C, so we just
need to plug these values into our table:
Remember: Hot water starts
with T = 80C
Heat
transferred
200 calories
T for Hot
Water
1C
New T for Hot Water
80C - 1C = 79C
Remember: Cold water starts
with T = 20C
T for Cold
Water
4C
Use new Hot Water T from
the line above = 79C
2000 calories
10C
79C - 10C = 69C
1C
69C - 1C = 68C
20C + 4C = 24C
Use new Cot Water T from the
line above = 24C
40C
Use new Hot Water T from
above = 69C
200 calories
New T for Cold water
24C + 40C = 64C
Use new Cot Water T from
above = 64C
4C
64C + 4C = 68C
And look: the New T Hot Water = New T Cold Water, so we are done.
Answer: Final temperature of the mixture = 68C.
Note: If you have gotten this answer with another method, that is fine.