Question

f you combine 320.0 mL of water at 25.00 °C and 100.0 mL of water at 95.00 °C, what is the final temperature of the mixture? Use 1.00 g/mL as the density of water.

Answer 1

The idea here is that the heat lost by the hot water sample will be equal to the heat absorbed by the room-temperature water sample.

−qlost=qabsorbed (*)

The minus sign is used here because heat lost carries a negative sign.

Your go-to equation here will be

q=m⋅c⋅ΔT, where

q - the amount of heat gained / lost
m - the mass of the sample
c - the specific heat of the substance
ΔT - the change in temperature, defined as the difference between the final temperature and the initial temperature

Since you're dealing with two samples of water, you don't need to know the value of water's specific heat to solve for the final temperature of the mixture, Tf.

So, the change in temperature for the two samples will be

For the hot sample: ΔThot=Tf−95.00∘C

For the warm sample: ΔTwarm=Tf−25.00∘C

If you take the density to be equal to 1.0 g mL−1, then the two volumes are equivalent to

320.0mL⋅1.0 g/1mL=320.0 g and 100.0mL⋅1.0 g/1mL=100.0 g

Use equation (*) top write

heat lost by hot sample=heat gained by warm sample

-100g.c _water.(Tf-95.0)^oC=320.0g.C_water.(Tf-25.0)^oC

This will get you

−100.0⋅Tf+9500∘C=320.0⋅Tf−8000^oC

220Tf=17500

Tf=79.54^oC