Question

Exactly 15.5 mL of water at 23.0 °C is added to a hot iron skillet. All of the water is converted to steam at 100.0°C. The mass of the skillet is 1.20 kg. What is the change in temperature of the skillet?

Answer 1

`\Delta T\approx 72.5^(\circ)C`

Step-by-step explanation:

Given:

Volume of water,
`V=15.5* 10^(-3)\ L=15.5* 10^(-3)\ m^3`

initial temperature of water,
`T_i=23^(\circ)C`

final temperature of water before forming steam,
`T_f=100^(\circ)C`

mass of hot skillet,
`m_s=1.2\ kg`

Usually skillets are made of cast iron and cast iron has a specific heat capacity of :

`c_s=460\ J.kg^(-1).K^(-1)`

Specific heat of water,
`c_w= 4186\ J.kg^(-1).K^(-1)`

Latent heat of vaporization of water,
`L_v=2.26* 10^6\ J.kg^(-1).K^(-1)`

Now, mass of water:

`\rm mass=density* volume`

`m_w=1000* 15.5* 10^(-3)`

`m_w=15.5* 10^(-3)\ kg`

Quantity of heat absorbed by the water of 23 degree Celsius to a point just before steaming:

`Q_w=m_w.c_w.(T_f-T_i)`

`Q_w=15.5* 10^(-3)* 4186* (100-23)`

`Q_w=4995.99\ J`

Quantity of heat absorbed by the water to vapourize:

`Q_v=m_w.L_v`

`Q_v=15.5* 10^(-3)* (2.26* 10^6)`

`Q_v=35030\ J`

∴Total heat lost by the skillet:

`Q=Q_w+Q_v`

`Q=40025.99\ J`

For change in temperature:

`Q=m_s.c_s.\Delta T`

`\Delta T=(Q)/(m_s.c_s)`

`\Delta T=(40025.99)/(1.2* 460)`

`\Delta T\approx 72.5^(\circ)C`