Answer:
The answer is below
Explanation:
A 21.0 g sample of aluminum, which has a specific heat capacity of 0.897 J g '°C ', is dropped into an insulated container containing 200.0 g of water at 25.0 °C and a constant pressure of 1 atm. The initial temperature of the aluminum is 90.1 °C Assuming no heat is absorbed from or by the container, or the surroundings, calculate the equilibrium temperature of the water. Be sure your answer has 3 significant digits.
Solution:
The amount of heat lost or gained by a substance is given by the formula:
Q = mcΔT
Where Q is the heat lost or gained, m is the mass of the substance, ΔT is the temperature change.
The law of conservation of energy states that energy cannot be created or destroyed but can be transferred.
Therefore the heat lost by aluminum = heat gained by water
[tex]Q_{lost}=Q_{gained}\\\\m_ac_a\Delta T_a=m_wc_w\Delta T_w\\\\m_a= mass \ of\ aluminium = 21g,c_a=specific\ heat\ capacity\ of\ aluminum\\=0.897\ J/g^oC, \Delta T_a=90.1^oC-x,x=equilibrum\ temperature\\m_w= mass \ of\ water = 200g,c_w=specific\ heat\ capacity\ of\ water\\=4.184\ J/g^oC, \Delta T_w=x-25^oC\\\\ hence:\\\\21*0.897*(90.1-x)=200*4.184*(x-25)\\\\1697.2137-18.837x=836.8x-20920\\\\838.8x+18.387=1697.2137+20920\\\\857.187x=22617.2137\\\\x=26.4^oC[/tex]
