A sample of iron is put into a calorimeter (see sketch at right) that contains of water. The iron sample starts off at and the temperature of the water starts off at . When the temperature of the water stops changing it's . The pressure remains constant at . Calculate the specific heat capacity of iron according to this experiment. Be sure your answer is rounded to the correct number of significant digits.

Therefore, the specific heat capacity of the iron is 0.567J/g.°C.

Note: The question is incomplete. The complete question is given as follows:

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.

Calculate the specific heat capacity of iron according to this experiment. Be sure your answer is rounded to the correct number of significant digits

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.