Question

If you combine 230.0 mL 230.0 mL of water at 25.00 ∘ C 25.00 ∘C and 120.0 mL 120.0 mL of water at 95.00 ∘ C, 95.00 ∘C, what is the final temperature of the mixture? Use 1.00 g/mL as the density of wate

Answer 1

Answer: The final temperature of the mixture is 49°C

Step-by-step explanation:

To calculate the mass of water, we use the equation:

`\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}`

• For cold water:

Density of cold water = 1 g/mL

Volume of cold water = 230.0 mL

Putting values in above equation, we get:

`1g/mL=\frac{\text{Mass of water}}{230.0mL}\\\\\text{Mass of water}=(1g/mL* 230.0mL)=230g`

• For hot water:

Density of hot water = 1 g/mL

Volume of hot water = 120.0 mL

Putting values in above equation, we get:

`1g/mL=\frac{\text{Mass of water}}{120.0mL}\\\\\text{Mass of water}=(1g/mL* 120.0mL)=120g`

When hot water is mixed with cold water, the amount of heat released by hot water will be equal to the amount of heat absorbed by cold water.

`Heat_{\text{absorbed}}=Heat_{\text{released}}`

The equation used to calculate heat released or absorbed follows:

`Q=m* c* \Delta T=m* c* (T_(final)-T_(initial))`

`m_1* c* (T_(final)-T_1)=-[m_2* c* (T_(final)-T_2)]` ......(1)

where,

q = heat absorbed or released

`m_1` = mass of hot water = 120 g

`m_2` = mass of cold water = 230 g

`T_(final)` = final temperature = ?°C

`T_1` = initial temperature of hot water = 95°C

`T_2` = initial temperature of cold water = 25°C

c = specific heat of water = 4.186 J/g°C

Putting values in equation 1, we get:

`120* 4.186* (T_(final)-95)=-[230* 4.186* (T_(final)-25)]`

`T_(final)=49^oC`

Hence, the final temperature of the mixture is 49°C