# Water is boiled in a container having a bottom of surface area 25 cm2, thickness 0.1 mm, and...

## Question:

Water is boiled in a container having a bottom of surface area 25 cm{eq}^2 {/eq}, thickness 0.1 mm, and thermal conductivity 50 W/m-C. 100 g of water is converted into steam per minute in the steady-state after the boiling starts. Assuming that no heat is lost to the atmosphere, calculate the temperature of the lower surface of the bottom. Latent heat of vaporization of water = {eq}2.26 \times 10^6 {/eq} J/kg.

## Heat Transfer:

The phenomenon in which the variation in the temperature of a body occurs is known as heat transfer. Generally, the heat transfer is indicated by Joule in the International standard of the unit system.

## Answer and Explanation:

**Given**

- The surface area of bottom of the container is {eq}A = 25\;{\rm{c}}{{\rm{m}}^{\rm{2}}} = 0.0025\;{{\rm{m}}^{\rm{2}}} {/eq}

- The thickness of the container is {eq}t = 0.1\;{\rm{mm}} = 0.0001\;{\rm{m}} {/eq}

- The thermal conductivity is {eq}k = 50\;{\rm{W/m^\circ C}} {/eq}

- The mass of the water is {eq}{m_w} = 100\;{\rm{g}} = 0.1\;kg {/eq}

- The time is {eq}T = \;1\;\min = 60\;{\rm{s}} {/eq}

- The latent heat of vaporization of water is {eq}{h_L} = 2.26 \times {10^6}\;{\rm{J/kg}} {/eq}

Note- The boiling temperature of water is {eq}{T_1} = 100{\rm{^\circ C}} {/eq}

The temperature of the lower surface of the bottom is calculated as,

{eq}\begin{align*} \dfrac{Q}{T} &= \dfrac{{kA\left( {T - {T_1}} \right)}}{t}\\ \dfrac{{{m_w}{h_L}}}{T} &= \dfrac{{kA\left( {T - {T_1}} \right)}}{t}\\ \dfrac{{\left( {0.1\;kg} \right)\left( {2.26 \times {{10}^6}\;{\rm{J/kg}}} \right)}}{{\left( {60\;{\rm{s}}} \right)}} &= \dfrac{{\left( {50\;{\rm{W/m^\circ C}}} \right)\left( {0.0025\;{{\rm{m}}^{\rm{2}}}} \right)\left( {T - 100^\circ {\rm{C}}} \right)}}{{\left( {0.0001\;{\rm{m}}} \right)}}\\ 1250\;{\rm{W/^\circ C}}\left( {T - 100^\circ {\rm{C}}} \right) &= 3766.67\;{\rm{J/s}}\\ T - 100^\circ {\rm{C}} &= 3.013^\circ {\rm{C}}\\ T &= 103.013{\rm{^\circ C}} \end{align*} {/eq}

**Thus, the temperature of the lower surface of the bottom is 103.013 degree Celsius.**

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from High School Physics: Help and Review

Chapter 17 / Lesson 12