50 g of an iron piece with initial temperature of 100C was dipped in 100g of water at 20C If the tempreatureof the mixture becmes 25.5C then calculate the specific heat of iron?

Answer 1

This is a classical question:

Assuming there is no heat loss/gain to/from surroundings:

Heat lost by iron = Heat gained by water

MCDO = mcdo

50 x C x (100 - 25.5) = 100 x 4.2 x (25.5 - 20)

The specific heat capacity of iron is C and that of water is 4.2 Jg-1K-1

Thus, C = 0.620 Jg-1K-1

Answer 2

We can use the principle of conservation of energy to solve this problem. The heat lost by the iron is equal to the heat gained by the water.

Let's denote the specific heat capacity of iron as (c_{\text{iron}}). Using the formula:

[\text{heat lost by iron} = \text{heat gained by water}]

[m_{\text{iron}}c_{\text{iron}}(T_f - T_{\text{final}}) = m_{\text{water}}c_{\text{water}}(T_{\text{final}} - T_{\text{initial}})]

[50 \times c_{\text{iron}}(25.5 - 25.5) = 100 \times 4.184 \times (25.5 - 20)]

Solve for (c_{\text{iron}}) to find the specific heat capacity of iron.