Question
Question: A piece of iron of mass \(200gm\) and temperature \(300{}^{o}C\) is dropped into \(1kg\) of water at...
A piece of iron of mass 200gm and temperature 300oC is dropped into 1kg of water at temperature 20oC. Predict the final equilibrium temperature of the water (Take c for iron as 450Jkg−1K−1)
Solution
When two bodies are at different temperatures, the heat flows from a hotter body to a colder body till equilibrium is attained. The heat absorbed or released is related to the mass, temperature change and specific heat of the substance. Substituting corresponding values in the relation, we can calculate the equilibrium temperature. Convert the units as required.
Formulas used:
mΔTQ=c
Complete answer:
The heat absorbed by 1gm of a substance when its temperature is increased by 1oC is called specific heat. Its SI unit is Jkg−1K−1. It is given by-
mΔTQ=c - (1)
Here, c is the specific heat
Q is the heat absorbed
m is the mass of the substance
ΔT is the change in temperature
Let the final temperature of the water be T, the specific heat for water is 4200Jkg−1K−1
. Water will absorb heat from the piece of iron
Applying eq (1) for water, we get,
Q=mcΔT⇒Q=1×4200(T−293)
The iron piece will give out heat, therefore, applying eq (1) for iron, we get,
Q′=200×10−3×450(573−T)
The heat released by iron is the heat absorbed by water, therefore,
Q=Q′⇒4200(T−20)=90(300−T)⇒4200T−84000=27000−90T⇒4290T=111000⇒T=4290111000⇒T=25.8oC∴T=25.8+273=298.8K
Therefore, the final equilibrium temperature of water is 298.8K.
Note:
When we have to calculate heat released or absorbed for a mole, we use molar specific heat. At equilibrium, the temperature of both the iron and the water is the same. The specific heat is highest for water. Specific heat is calculated at a constant volume. When temperature is constant, heat changes result in the change of volume. The specific heat is a constant quantity which depends on the substance and temperature.