Question

Suppose someone pours 0.250 kg of 20.0ºC water (about a cup) into a 0.500-kg aluminum pan with a temperature of 150ºC. Assume that the pan is then placed on an insulated pad and that a negligible amount of water boils off. What is the temperature when the water and pan reach thermal equilibriuma short time later?

Answer 1

Answer : The temperature when the water and pan reach thermal equilibrium short time later is,
`59.10^oC`

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

`q_1=-q_2`

`m_1* c_1* (T_f-T_1)=-m_2* c_2* (T_f-T_2)`

where,

`c_1` = specific heat of aluminium =
`0.90J/g^oC`

`c_2` = specific heat of water =
`4.184J/g^oC`

`m_1` = mass of aluminum = 0.500 kg = 500 g

`m_2` = mass of water = 0.250 kg = 250 g

`T_f` = final temperature of mixture = ?

`T_1` = initial temperature of aluminum =
`150^oC`

`T_2` = initial temperature of water =
`20^oC`

Now put all the given values in the above formula, we get:

`500g* 0.90J/g^oC* (T_f-150)^oC=-250g* 4.184J/g^oC* (T_f-20)^oC`

`T_f=59.10^oC`

Therefore, the temperature when the water and pan reach thermal equilibrium short time later is,
`59.10^oC`