Calculate the final temperature when 50ml of water at 80 degree Celsius are added to 25 ml of water at 25 degree Celsius?

Answer 1

Okay, so we have some amount of water that is hot, and some amount of water that is cold. Mix em together and you will get something warm. At the end, both the hot water and cold water you started with will be the same temperature. If both will equal the same temperature at the end of the reaction, we can make them equal to each other and solve for "X".

s = specific heat (specific heat of water is 4.18 J/ degree Celsius x gram)

m = mass

(Tf - Ti) aka. "delta T" = change in temperature, Final - Initial.

However since one of them is losing heat and the other is gaining heat energy, the equation is:

qhot = - qcold

s1 x m1 x (Tf - Ti) = - s2 x m2 x (Tf - Ti)

(4.18 J / C x g) x (50.g) x (X - 60 C) = - (4.18 J / C x g) x (25g) x (X - 20 C)

DROP UNITS and solve for "X"

depending on whether you round or not, I'll put down exact numbers and round at end

209X - 12540 = - 104.5X + 2090

313.5X = 14630

X = 46.66666

In which case you can round it to 47 C.

So to sum it up, if you mix these two together, the final temperature will be 47 degrees Celsius.

Answer 2

50 g x (60 - Tf) = 25 g x (Tf - 25)

3000 -50Tf = 25Tf - 625

3625 = 75Tf

Tf = 48.3 deg C (48 deg)

NOTE: I have assumed that 1 mL of water = 1 gram at 60 deg C and at 25 deg C. Obviously that isn't so. The differences won't significantly impact the answer.

Answer 3

Let the final temperature attained be toC.

Assuming that there is no dissipation of heat to the surroundings, nor is there any external source of gaining heat, we can equate the amount of total heat loss to the amount of total heat gain. So,

250*1*(t-25)=130*1*(95-t)

or, 250t - 6250 = 12350 - 130t

or, 250t + 130t = 12350 +6250

or, 380t = 18600

or, t = 18600 / 380

ot, t = 48.95

So, final temperature is 48.95oC

Answer 4

If you ignore the thermal expansion of water, then the resulting mixture will be at a temperature of 61+2/3 = 61.7 deg C.