Respuesta :
Answer : The final temperature is, [tex]337.8K[/tex]
Explanation :
[tex]Q_{absorbed}=Q_{released}[/tex]
As we know that,
[tex]Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})[/tex]
[tex]m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)][/tex] .................(1)
where,
q = heat absorbed or released
[tex]m_1[/tex] = mass of water at [tex]90^oC[/tex] = 150 g
[tex]m_2[/tex] = mass of water at [tex]30^oC[/tex]= 100 g
[tex]T_{final}[/tex] = final temperature = ?
[tex]T_1[/tex] = temperature of lead = [tex]90^oC=273+90=363K[/tex]
[tex]T_2[/tex] = temperature of water = [tex]30^oC=273+30=303K[/tex]
[tex]c_1\text{ and }c_2[/tex] = same (for water)
Now put all the given values in equation (1), we get
[tex]150\times (T_{final}-363)=-[100\times (T_{final}-303)][/tex]
[tex]T_{final}=337.8K[/tex]
Therefore, the final temperature is, [tex]337.8K[/tex]
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.
[tex]Heat_{\text{absorbed}}=Heat_{\text{released}}[/tex]
The equation used to calculate heat released or absorbed follows:
[tex]Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})[/tex]
[tex]m_1\times c\times (T_{final}-T_1)=-[m_2\times c\times (T_{final}-T_2)][/tex] ......(1)
where,
q = heat absorbed or released
[tex]m_1[/tex] = mass of hot water = 150.0 g
[tex]m_2[/tex] = mass of cold water = 100.0 g
[tex]T_{final}[/tex] = final temperature = ?°C
[tex]T_1[/tex] = initial temperature of hot water = 90.0°C
[tex]T_2[/tex] = 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:
[tex]150\times 4.186\times (T_{final}-90)=-[100\times 4.186\times (T_{final}-30)]\\\\T_{final}=66^oC[/tex]
Hence, the final temperature of the mixture is 66°C
