Question

What mass of water at 23.0°C must be allowed to come to thermal equilibrium with a 1.75-kg cube of aluminum initially at 150°C to lower the temperature of the aluminum to 68.0°C? Assume any water turned to steam subsequently recondenses.

Answer 1

`m = 0.686 kg`

Step-by-step explanation:

For thermal equilibrium we know that

Heat given by aluminium = Heat absorbed by water

For water

`Q_1 = m_1 s_1\Delta T_1`

`Q_1 = m(4186)(68 - 23)`

For aluminium

`Q_2 = m_2s_2\Delta T_2`

`Q_2 = (1.75)(900)(150 - 68)`

`Q_2 = 129150 J`

now by condition mentioned above we have

`m(4186)(68 - 23) = 129150`

`m = 0.686 kg`

Answer 2

0.73kg

Step-by-step explanation:

let,

The mass of the water as
`m_(water)`

Given:

Mass of the aluminum,
`m_(Al) =1.85 kg`

Initial temperature of the water,
`T_1=23^oC`

Initial temperature of the aluminum,
`T_2=150^oC`

The final temperature of the aluminum,
`T_3=68^oC`

Now,

the heat gained by the water = the heat lost by the aluminium

`m_w* C_w*(T_3-T_1)=m_(Al)* C_(Al)* (T2-T3)`

where,

`C_w \ and\ C_(Al)` = specific heat of water and aluminium

`C_w = 4186\ J/kg^oC \ and\ C_(Al)=904 J/kg^oC`

substituting the values in the equation, we get

`m_w* 4186*(68-23)=1.75* 904* (150-63)`

thus,

`m_w=0.73kg`

Hence, the required mass of the water required is 0.73kg