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aev [14]
3 years ago
12

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 t

o lower the temperature of the aluminum to 68.0°C? Assume any water turned to steam subsequently recondenses.
Physics
2 answers:
garri49 [273]3 years ago
4 0

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

Ahat [919]3 years ago
4 0

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

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