40g water at10c are mixed with 25g of water at 90c what is the equilibrium temperature?

Answer 1

Choose something convenient (like 0°C) as a reference. Then you have 40g of water containing 400 Calories of heat energy and 25g of water containing 2,250 calories of heat energy. That's 65g of water containing 2,650 calories so the temperature is 2,650/65 = 40.77 °C above the reference (which we cleverly chose to be zero ☺) so the final equilibrium temperature is 40.77°C.

These kinds of problems are a *lot* more fun when they involve phase transitions and/or different materials with different heat capacities ☺


Doug

Answer 2

Equilibrium means Q1 + Q2 = 0.

Therefore,

mass 1 * specific heat 1 * (Tf - Ti1) + mass 2 * specific heat 2 * (Tf - Tf1) = 0

Use equation:

m1 c1 (Tf - Ti1) + m2 c2 (Tf - Ti2) = 0

Given:

m1 = 40 g = 0.040 kg
m2 = 25 g = 0.025 kg
Ti1 = 10 C = 283.15 K
Ti2 = 90 C = 363.15 K
c of water = 4186 J/(kg*K) = c1 = c2 (Only one substance)

Substitute:

(0.040)(4186)(Tf - 283.15) + (0.025)(4186)(Tf - 363.15) = 0

Solve for Tf...

Tf = 313.919 K = 40.769 C

Answer 3

40*10 + 25*90 = 2650
2650/65 = approx 40.8degC