Answer:
Therefore, the specific heat capacity of the iron is 0.567J/g.°C.
<em>Note: The question is incomplete. The complete question is given as follows:</em>
<em>A 59.1 g sample of iron is put into a calorimeter (see sketch attached) that contains 100.0 g of water. The iron sample starts off at 85.0 °C and the temperature of the water starts off at 23.0 °C. When the temperature of the water stops changing it's 27.6 °C. The pressure remains constant at 1 atm.
</em>
<em>
Calculate the specific heat capacity of iron according to this experiment. Be sure your answer is rounded to the correct number of significant digits</em>
Explanation:
Using the formula of heat, Q = mc∆T
where Q = heat energy (Joules, J), m = mass of a substance (g)
c = specific heat capacity (J/g∙°C), ∆T = change in temperature (°C)
When the hot iron is placed in the water, the temperature of the iron and water attains equilibrium when the temperature stops changing at 27.6 °C. Since it is assumed that heat exchange occurs only between the iron metal and water; Heat lost by Iron = Heat gained by water
mass of iron = 59.1 g, c = ?, Tinitial = 85.0 °C, Tfinal = 27.6 °C
∆T = 85.0 °C - 27.6 °C = 57.4 °C
mass of water = 100.0 g, c = 4.184 J/g∙°C, Tinitial = 23.0 °C, Tfinal = 27.6 °C
∆T = 27.6°C - 23.0°C = 4.6 °C
Substituting the values above in the equation; Heat lost by Iron = Heat gained by water
59.1 g * c * 57.4 °C = 100.0 g * 4.184 J/g.°C * 4.6 °C
c = 0.567 J/g.°C
Therefore, the specific heat capacity of the iron is 0.567 J/g.°C.
= 0.800 mol Al