Calculate the final temperature when 50ml of water at 80 degree Celsius are added to 25 ml of water at 25 degree Celsius?
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 = - qcolds1 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 end209X - 12540 = - 104.5X + 2090313.5X = 14630X = 46.66666In 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.