if a 100g sample of water at 60.0 C is added to a 100.0 g sample of water at 10.0 C, determine the final temperature of the water. Assume no heat is lost to the surroundings

Answer 1

The equation to be used for the problem is mcΔT = mcΔT.
From the given problem, we need to determine whether the heat is absorbed or evolved. When heat is absorbed, the sign is positive, and when heat is evolved, it is negative; therefore +q = -q, due to the reason then when the 100g of sample at 60.0C was added, it absorbed heat while the 100.0g sample of water at 10.0C evolved the heat towards the previous sample.

Applying the equation with the proper signs, we get mcΔT = -mcΔT, the specific heat of water is cancelled out right away.

Required: Final Temperature (T2)
mcΔT = (1)
-mcΔT = (2)

(1) m = 100.00g
T2-60.0C
(2) m = 100.00g
T2-10.0C

(100g) (T2-60.0C) = -(100g) (T2-10.0C)
Multiply the mass with the (T2-T1) equation.
100gT2 - 6000C = -100gT2 - 1000C
Transpose the values.
100gT2+100gT2 = 6000C - 1000C
200gT2 = 5000C
T2 = 5000C/200g
T2 = 25C.

Answer 2

62.5