Respuesta :
Answer:
[tex]\Delta T\approx 72.5^{\circ}C[/tex]
Explanation:
Given:
Volume of water, [tex]V=15.5\times 10^{-3}\ L=15.5\times 10^{-3}\ m^3[/tex]
initial temperature of water, [tex]T_i=23^{\circ}C[/tex]
final temperature of water before forming steam, [tex]T_f=100^{\circ}C[/tex]
mass of hot skillet, [tex]m_s=1.2\ kg[/tex]
Usually skillets are made of cast iron and cast iron has a specific heat capacity of :
[tex]c_s=460\ J.kg^{-1}.K^{-1}[/tex]
Specific heat of water, [tex]c_w= 4186\ J.kg^{-1}.K^{-1}[/tex]
Latent heat of vaporization of water, [tex]L_v=2.26\times 10^6\ J.kg^{-1}.K^{-1}[/tex]
Now, mass of water:
[tex]\rm mass=density\times volume[/tex]
[tex]m_w=1000\times 15.5\times 10^{-3}[/tex]
[tex]m_w=15.5\times 10^{-3}\ kg[/tex]
Quantity of heat absorbed by the water of 23 degree Celsius to a point just before steaming:
[tex]Q_w=m_w.c_w.(T_f-T_i)[/tex]
[tex]Q_w=15.5\times 10^{-3}\times 4186\times (100-23)[/tex]
[tex]Q_w=4995.99\ J[/tex]
Quantity of heat absorbed by the water to vapourize:
[tex]Q_v=m_w.L_v[/tex]
[tex]Q_v=15.5\times 10^{-3}\times (2.26\times 10^6)[/tex]
[tex]Q_v=35030\ J[/tex]
∴Total heat lost by the skillet:
[tex]Q=Q_w+Q_v[/tex]
[tex]Q=40025.99\ J[/tex]
For change in temperature:
[tex]Q=m_s.c_s.\Delta T[/tex]
[tex]\Delta T=\frac{Q}{m_s.c_s}[/tex]
[tex]\Delta T=\frac{40025.99}{1.2\times 460}[/tex]
[tex]\Delta T\approx 72.5^{\circ}C[/tex]
