Respuesta :
Answer: The final temperature of the system will be [tex]13.14^0C[/tex]
Explanation:
[tex]heat_{absorbed}=heat_{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 steam = 25 g
[tex]m_2[/tex] = mass of water = 0.2384 kg = 238.4 g (1kg=1000g)
[tex]T_{final}[/tex] = final temperature = ?
[tex]T_1[/tex] = temperature of steam = [tex]116^oC[/tex]
[tex]T_2[/tex] = temperature of water = [tex]8^oC[/tex]
[tex]c_1[/tex] = specific heat of steam = [tex]1.996J/g^0C[/tex]
[tex]c_2[/tex] = specific heat of water= [tex]4.184J/g^0C[/tex]
Now put all the given values in equation (1), we get
[tex]25g\times 1.996J/g^0C\times (T_{final}-116)=-[238.4g\times 4.184J/g^0C\times (T_{final}-8)][/tex]
[tex]T_{final}=13.14^0C[/tex]
Therefore, the final temperature of the system will be [tex]13.14^0C[/tex]
