Final Temperature?

Question

if 50mL of 10Celcius water is added to 40 ml of 65 Celsius, calculate the final temperature of the mixture assuming no heat is lost to the surroundings including the container.

Answer 1

The solution is to calculate the total heat contained by the mixture, and then divide by the total mass of water:

1. Calculate the heat in the 50 ml of 10 C water.

h1 = 1 cal/g C * 50 g * 10 deg C = 500 cal

2. Calculate heat in the 40 ml of 65 C water

h1 = 1 cal/g C * 40 g * 65 deg C = 2600 cal

3. Calculate the total heat, divide by the mass

Total heat = 3100 cal

3100 cal / 90 g = 34.4 deg C

Answer 2

27.5'C ? in basic terms by way of fact if it have been 100g of each and every and each temp itd be 30'C, yet by way of fact its a million / 4 lots much less of 10'C it is going to minus 2.5 'C if ya get me so: 100g of 20'C + 100g of 40'C = 200g of 30'C ?? agreed? (if im now no longer acceptable then some component else wont be) yet its now no longer 100g of 40'C, its 75g, it has lost a million / 4 of its weight. and by way of fact that the nice and comfortable water is now lots much less, the acceptable water mixture would be less warm than if the two parts of water have been an comparable. by way of fact the weights arnt even, there is greater desirable chilly water. see how i artwork that? :)

Answer 3

-q(g) +q(L) = 0

-50T2 +500 -40T2 +2600 = 0
90T2 = 3100
T2 = 34.4C

Answer 4

34.4º C