Question

A coffee cup calorimeter contains 152.18 g of water at 20.90 °C. A 55.336 g piece of iron is heated to 98.37 °C. The piece of iron is added to the coffee cup caloriemter and the contents reach thermal equilibrium at 23.60 °C. The specific heat capacity of iron is 0.449 J g⋅K and the specific heat capacity of water is 4.184 J g⋅K . How much heat, q , is lost by the piece of iron?

How much heat, q, is gained by the water?

The difference between the heat lost by the piece of iron and the heat gained by the water is due to heat transfer to the styrofoam and the heat required to raise the temperature of the calorimeter. What is the heat capacity of the styrofoam calorimeter in joules per kelvin (J/K)?

What would be the final temperature of the system if all of the heat lost by the iron was absorbed by the water?

Answer 1

q_iron = -1857.7J

q_water = 1719.1 J

C_calorimeter = 51.33 J/K

T_final = 23.81 ^oC

Explanation

mass of iron = 55.336 g

initial temperature of iron = 98.37 ^oC

final temperature of iron = 23.60 ^oC

heat lost by iron, q_iron = (mass of iron) * (specific heat of iron) * (final temperature of iron - initial temperature of iron)

q_iron = (55.336 g) * (0.449 J/g.^oC) * (23.60 ^oC - 98.37 ^oC)

q_iron = -1857.7 J

Heat gained by water, q_water = (mass of water) * (specific heat of water) * (final temperature of water - initial temperature of water)

q_water = (152.18 g) * (4.184 J/g.^oC) * (23.60 ^oC - 20.90 ^oC)

q_water = 1719.1 J

Since heat energy is conserved, q_iron + q_water + q_calorimeter = 0

-1857.7 J + 1719.1 J + q_calorimeter = 0

q_calorimeter = 1857.7 J - 1719.1 J

q_calorimeter = 138.6 J

C_calorimeter = (q_calorimeter) / (final temperature of water - initial temperature of water)

C_calorimeter = (138.6 J) / (23.60 ^oC - 20.90 ^oC)

C_calorimeter = 51.33 J/K