Question

A hot copper pan is dropped into at tub of hot water. If the water's temperature rises, what happens to the temperature of the pan? How will you know when the water and the copper pan reach thermal equilibrium?

Answer 1

Explanation:

As it is given that hot pan of copper is dropped into a tub of hot water and the temperature of water rises. This means that heat from the pan has been released and this heat is gained by water.

As a result, temperature of copper pan has decreased and this decrease will continue till the time temperature of both copper pan and water will reach the same temperature.

As thermal energy is defined as the energy in which when two objects come in physical contact with each other then no exchange of heat energy will take place.

Thus, we can conclude that when temperature of both copper pan and water will be equal then it means that both of them has reached thermal equilibrium.

Answer 2

Answer:

• Thus the overall temperature of the water increases and the temperature of the pan decreases if the water is less hotter than the pan and vice-versa if the water is hotter than the pan.

$m_c.c_c.(T_f-T_{ic})=m_w.c_w.(T_f-T_{iw})$

Explanation:

• Thus the overall temperature of the water increases and the temperature of the pan decreases if the water is less hotter than the pan.

When a hot copper pan is dropped into a tub of hot water then the water temperature further rises until it reaches the boiling point (ideally).

But what happens practically is as if the pan is too hot and comes in contact with the hot water suddenly then the water molecules near to it firstly gain the heat from it and if their rate of heat absorption is faster then they convert into steam and the bubble is observed in the tub water which may rise up but due to relatively cooler water in contact it condensed back within the liquid. Only the surface water open to atmosphere may directly vaporize.

• When the water is hotter than the pan then the pan will absorb the heat to increase its temperature and the water will cool down.

When the water and copper pan reach the thermal equilibrium can be precisely known only by the calculation where the following quantities must be known:

(heat lost by one body is the heat gained by the other body)

• Heat transfer through copper = Heat transfer through water

$Q_c=Q_w$

$m_c.c_c.(T_f-T_{ic})=m_w.c_w.(T_f-T_{iw})$

where:

$T_f=$ final temperature of the system

$m_c\ \&\ m_w=$ mass of copper and water respectively

$c_c\ \&\ c_w=$ specific heat of copper and water respectively

$T_{ic}\ \&\ T_{iw}=$ initial temperature of copper and water respectively