Question

If you combine 300.0 mL of water at 25.00 ∘C and 140.0 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

`T_F=47.3\°C`

Step-by-step explanation:

Hello!

In this case, since we have hot and cold water, we infer that as hot water cools down, cool water heats up based on the first law of thermodynamics; thus, we can write:

`Q_(hot)+Q_(cold)=0`

In such a way, we can write the expression in terms of mass, specific heat and temperature change:

`m_(hot)C_(hot)(T_F-T_(hot))+m_(cold)C_(cold)(T_F-T_(cold))=0`

However, since they both have the same specific heat and the same mL are in g due to the 1.00-g/mL density, we obtain:

`m_(hot)(T_F-T_(hot))+m_(cold)(T_F-T_(cold))=0\\\\T_F=(m_(hot)T_(hot)+m_(cold)T_(cold))/(m_(hot)+m_(cold))`

Now, we plug in to obtain:

`T_F=(140.0g*95.00\°C+300g*25.00\°C)/(140.0g+300g)\\\\T_F=47.3\°C`

Best regards!