Respuesta :
Answer : The final temperature of water is, [tex]119.84^oC[/tex]
Solution :
[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_g\times (T_{final}-T_2)=-[m_2\times c_w\times (T_{final}-T_1)][/tex] .................(1)
where,
[tex]m_1[/tex] = mass of glass = 32.50 g
[tex]m_2[/tex] = mass of water = 57 g
[tex]T_{final}[/tex] = final temperature of water and glass = [tex]119.2^oC[/tex]
[tex]T_2[/tex] = initial temperature of water = ?
[tex]T_1[/tex] = initial temperature glass = [tex]115^oC[/tex]
[tex]c_w[/tex] = specific heat of water = [tex]4.184J/g^oC[/tex]
[tex]c_g[/tex] = specific heat of glass = [tex]0.840J/g^oC[/tex]
Now put all the given values in equation (1), we get
[tex]m_1\times c_g\times (T_{final}-T_2)=-[m_2\times c_w\times (T_{final}-T_1)][/tex]
[tex](32.50g)\times (0.840J/g^oC)\times (119.2^oC-115^oC)=-[(57g)\times (4.184J/g^oC)\times (119.2^oC-T_1)][/tex]
[tex]T_1=119.84^oC[/tex]
Therefore, the final temperature of water is, [tex]119.84^oC[/tex]
