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

Answer:

$m = 0.686 kg$

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

Answer:

0.73kg

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

$m_w\times C_w\times(T_3-T_1)=m_{Al}\times C_{Al}\times (T2-T3)$

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

$m_w\times 4186\times(68-23)=1.75\times 904\times (150-63)$

thus,

$m_w=0.73kg$

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