Answer:
[tex]T_f=5.0116^{\circ}C[/tex]
Explanation:
Given:
- mass of water, [tex]m_w=0.25\ kg[/tex]
- initial temperature of water, [tex]T_i_w=20^{\circ}C[/tex]
- initial temperature of pan, [tex]T_i_p=173^{\circ}C[/tex]
- mass of pan, [tex]m_p=0.6\ kg[/tex]
- mass of water evapourated, [tex]m_v=0.03\ kg[/tex]
- specific heat of water, [tex]c_w=4186\ J.kg^{-1}.K^{-1}[/tex]
- specific heat of aluminium pan, [tex]c_a=900\ J.kg^{-1}.K^{-1}[/tex]
- latent heat of vapourization, [tex]L=2256000\ J.kg^{-1}[/tex]
Using the equation of heat:
Here, initially certain mass of water is vapourised first and then the remaining mass of water comes in thermal equilibrium with the pan.
[tex]m_p.c_a.(T_{ip}-T_f)=m_v.L+(m_w-m_v).c_w.(T_f-T_{iw})[/tex]
[tex]0.6\times 900\times (173-T_f)=0.03\times 2256000+(0.25-0.03)\times 4186\times (T_f-20)[/tex]
[tex]T_f=5.0116^{\circ}C[/tex]