a 5.00g sample of aluminum pellets(specific heat capacity=0.89J/degreesCxg) and a 10.00g sample of iron pellets(specific heat capacity=0.45J/degreesCxg) are heated to 100.0degreesC. The mixture of hot iron and aluminum is then dropped into 97.3g water at 22.0degrees C. Calculate the final temperature of the metal and water mixture, assuming no heat loss to the surroundings.
Answers
Given:
Specific Heat Capacity of Iron = 0.45 J °C-1 g-1 => 450 J °C-1 kg-1
Specific Heat Capacity of Al = 0.89 J °C-1 g-1 => 890 J °C-1 kg-1
=>mass of aluminium = 0.005 kg
=>mass of Iron = 0.01 kg
mass of water = 0.0973 kg
Initial Temperature of Al and Fe = 100°C
Initial Temperature of Water = 22°C
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Formula Used:
Q = m * c * △T
Where,
Q = Heat Absorbed Or Given Out
m = mass of body
c = specific heat capacity of body
△T = Change in temperature
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=> So, 5 grams of Aluminium and 10 grams of Iron pellets are heated to 100°C and added to 97.3 grams of water.
=> Final temperature = ?
=>All the heat given out by the aluminium and iron is absorbed by the water.
=> Let final Temperature be T.
For Heat Gained by Water:
T1 = 22 °C
T = ?
△T = T-22 °C
m = 0.0973 kg
c = 4200 J kg-1 °C-1
Q = 0.0973 kg * 4200 J kg-1 °C-1 * (T-22) °C
Q1 = 408.66 * (T-22) J
For Heat Lost By Aluminium:
T1 = 100 °C
T = ?
△T = 100-T °C
m = 0.005 kg
c = 890 J kg-1 °C-1
Q = 0.005 kg * 890 J kg-1 °C-1 * (100-T) °C
Q2 = 4.45 * (100-T) J
For Heat Lost By Iron:
T1 = 100 °C
T = ?
△T = 100-T °C
m = 0.01 kg
c = 450 J kg-1 °C-1
Q = 0.01 kg * 450 J kg-1 °C-1 * (100-T) °C
Q3 = 4.5 * (100-T) J
Now,
Q1 = Q2 + Q3
408.66 * (T-22) = 4.45 * (100-T) + 4.5 * (100-T)
=> 408.66T - 8990.52 = 445 - 4.45T + 450 - 4.5T
=> 408.66T - 8990.52 = 895 - 8.95T
=> 408.66T + 8.95T = 895 + 8990.52
=> 417.61 T = 9885.52
=> T = 9885.52/417.61
=> T = 23.67 °C