Question

When 165 mL of water at 22°C is mixed with 85 mL of water at 82°C, what is the final temperature? (Assume that no heat is lost to the surroundings; d of water = 1.00 g/mL, specific heat capacity of water is 4.184 J/g.K)

A. 42°C
B. 52°C
C. 61°C
D. -61°C
E. None of these is within 5% of the correct answer

Answer 1

The final temperature when two quantities of water are mixed can be found using the conservation of energy and the equation: 165g * 4.184 J/g.K * (T_f - 22°C) = -85g * 4.184 J/g.K * (T_f - 82°C), solving for T_f gives the final temperature.

Step-by-step explanation:

To determine the final temperature when two quantities of water at different temperatures are mixed, we can use the principle of conservation of energy. According to this principle, heat lost by the warmer water will equal the heat gained by the cooler water, assuming no heat is lost to the surroundings. The specific heat capacity of water and the amounts of water at each temperature are needed for this calculation.

Considering the density of water is 1.00 g/mL, we have 165 mL of water at 22°C weighing 165 grams and 85 mL of water at 82°C weighing 85 grams. Applying the formula Q = mcΔT (where Q is heat in joules, m is mass in grams, c is the specific heat capacity in J/g.°K, and ΔT is the change in temperature in °C), we set up the equation for heat lost and gained:
m1 * c * (ΔT1) = -m2 * c * (ΔT2)

Let T_f be the final temperature of the mixture, then:

• 165g * 4.184 J/g.K * (T_f - 22°C) = -85g * 4.184 J/g.K * (T_f - 82°C)

Solving for T_f, we get the final temperature of the water mixture.