A container contains 200g of water at initial temperature of 30°C. An iron nail of mass 200g at temperature of 50°C is immersed in the water. What is the final water temperature? State the assumptions you need to make in your calculations.
[Given the value of specific heat capacity of water is 4200 J kg^-1 °C^-1 and that of iron is
450 J kg^-1 °C^-1]
Question
Answer:
The final temperature is 31.94°
Explanation:
The mass of the water in the container m₁ = 200 g = 0.2 kg
The initial temperature of the water, T₁₁ = 30°C
The mass of the iron, m₂ = 200 g = 0.2 kg
The temperature of the iron T₂₁= 50°C is immersed in the water,
The specific heat capacity of the water, c₁ = 4200 J/(kg·°C)
The specific heat capacity of the iron, c₂ = 450 J/(kg·°C)
Heat capacity relation is given by the formula;
Heat capacity Q = Mass, m × Specific heat capacity, c × Temperature change, (T₂ – T₁)
Given that energy can neither be created nor destroyed, and with the assumption that all the heat lost by the nail is gained by the water we have;
Heat lost by iron nail = Heat gained by the water
m₁ × c₁ × (T₂ – T₁₁) = m₂ × c₂ × (T₂₁ – T₂)
Where, T₂ is the final temperature
0.2 kg × 4200 J/(kg·°C) × (T₂ – 30) = 0.2 kg × 450 J/(kg·°C) × (50° – T₂)
840·T₂ – 25200 = 4500 – 90·T₂
4500 + 25200 = 840·T₂ + 90·T₂
29700 = 930·T₂
T₂ = 29700/930 = 31.94°.
The final temperature = 31.94°.