Question

A piece of iron of mass 200g and tempreture 300°C is dropped into 1.00 kg of water of tempreture 20°C. Predict the final equilibrium of the water.

Answer 1

Answer:

The final equilibrium T_{f} = 25.7[°C]

Explanation:

In order to solve this problem we must have a clear concept of heat transfer. Heat transfer is defined as the transmission of heat from one body that is at a higher temperature to another at a lower temperature.

That is to say for this case the heat is transferred from the iron to the water, the temperature of the water will increase, while the temperature of the iron will decrease. At the end of the process a thermal balance is found, i.e. the temperature of iron and water will be equal.

The temperature of thermal equilibrium will be T_f.

The heat absorbed by water will be equal to the heat rejected by Iron.

$Q_{iron} = Q_{water}$

Heat transfer can be found by means of the following equation.

$Q_{iron}=m*C_{piron}*(T_{i}-T_{f})$

where:

Qiron = Iron heat transfer [kJ]

m = iron mass = 200 [g] = 0.2 [kg]

T_i = Initial temperature of the iron = 300 [°C]

T_f = final temperature [°C]

$Q_{water}=m*C_{pwater}*(T_{f}-T_{iwater})$

Cp_iron = 437 [J/kg*°C]

Cp_water = 4200 [J/kg*°C]

$0.2*437*(300-T_{f})=1*4200*(T_{f}-20)\\26220-87.4*T_{f}=4200*T_{f}-84000\\26220+84000=4200*T_{f}+87.4*T_{f}\\110220 = 4287.4*T_{f}\\T_{f}=25.7[C]$