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 is, $337.8K$

Explanation :

$Q_{absorbed}=Q_{released}$

As we know that,  

$Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})$

$m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]$         .................(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

$150\times (T_{final}-363)=-[100\times (T_{final}-303)]$

$T_{final}=337.8K$

Therefore, the final temperature is, $337.8K$

Answer 2

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

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:

$Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})$

$m_1\times c\times (T_{final}-T_1)=-[m_2\times c\times (T_{final}-T_2)]$      ......(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:

$150\times 4.186\times (T_{final}-90)=-[100\times 4.186\times (T_{final}-30)]\\\\T_{final}=66^oC$

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