Question

What is the final temperature when 150.0 g of water at 90.0 °c is added to 100.0 g of water at 30.0 OC? Note that C, of water cancels. -qhot

Answer 1

Answer: The final temperature of the mixture is 66°C

Step-by-step explanation:

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 = 150.0 g

`m_2` = mass of cold water = 100.0 g

`T_(final)` = final temperature = ?°C

`T_1` = initial temperature of hot water = 90.0°C

`T_2` = initial temperature of cold water = 30.0°C

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

Putting values in equation 1, we get:

`150* 4.186* (T_(final)-90)=-[100* 4.186* (T_(final)-30)]\\\\T_(final)=66^oC`

Hence, the final temperature of the mixture is 66°C

Answer 2

Answer : The final temperature is,
`337.8K`

Explanation :

`Q_(absorbed)=Q_(released)`

As we know that,

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

`m_1* c_1* (T_(final)-T_1)=-[m_2* c_2* (T_(final)-T_2)]` .................(1)

where,

q = heat absorbed or released

`m_1` = mass of water at
`90^oC` = 150 g

`m_2` = mass of water at
`30^oC`= 100 g

`T_(final)` = final temperature = ?

`T_1` = temperature of lead =
`90^oC=273+90=363K`

`T_2` = temperature of water =
`30^oC=273+30=303K`

`c_1\text{ and }c_2` = same (for water)

Now put all the given values in equation (1), we get

`150* (T_(final)-363)=-[100* (T_(final)-303)]`

`T_(final)=337.8K`

Therefore, the final temperature is,
`337.8K`