Please show each step: Consider a mixture of 100 grams of water at

25 ° C with 75 grams of water at 40 ° C. Calculate the final temperature of the sample.

the warmer water looses heat, while the cooler water gains heat

let t equal the final temperature

100 (t - 25) = 75 (40 - t)

solve for t

@Scott

I'm not too good at this, would it be 31° C?

To calculate the final temperature of the mixture, we need to use the concept of heat transfer. The heat gained by the colder water will be equal to the heat lost by the hotter water. Generally, we use the equation:

Q = mcΔT

Where:
Q = Heat transferred
m = Mass of the substance
c = Specific heat capacity
ΔT = Change in temperature

In this case, we have two masses of water at different temperatures. Let's label them as water 1 (100 grams at 25 °C) and water 2 (75 grams at 40 °C).

Step 1: Calculate the heat lost by water 1:
Q1 = m1 * c * ΔT1
= 100 g * (specific heat capacity of water) * (final temperature - 25 °C)

Step 2: Calculate the heat gained by water 2:
Q2 = m2 * c * ΔT2
= 75 g * (specific heat capacity of water) * (40 °C - final temperature)

Step 3: Set the heat lost equal to the heat gained:
Q1 = Q2

Step 4: Solve for the final temperature:
100 g * (specific heat capacity of water) * (final temperature - 25 °C) = 75 g * (specific heat capacity of water) * (40 °C - final temperature)

Simplifying the equation, we get:
100 * (final temperature - 25) = 75 * (40 - final temperature)

Step 5: Solve for the final temperature:
100 * final temperature - 2500 = 75 * 40 - 75 * final temperature

Combine like terms:
100 * final temperature + 75 * final temperature = 75 * 40 + 2500
175 * final temperature = 3000 + 2500
175 * final temperature = 5500

Step 6: Divide both sides by 175 to solve for the final temperature:
final temperature = 5500 / 175
final temperature ≈ 31.43°C

Therefore, the final temperature of the mixture will be approximately 31.43°C.