Question

4) A 100g piece of iron at 150°C is emersed in 268.5 g of water at 20°C. the temperature of both iron and water became 25°C. If the specific heat of iron is 0.449 J/g.K, find the specific heat of water?

Answer 1

The specific heat of water would be 4.18 J/gK.

Step-by-step explanation:

You can realize that iron is losing heat and water is gaining heat. Based on the energy conservation law:

`Heat\text{ lost by iron = Heat gained by water.}`

We want to find the specific heat of water 'c'. The formula of each heat - lost or gained- is:

`q=c\cdot m\cdot\Delta T.`

You can see that the change of temperature of iron is from 150 °C to 25 °C and the change of temperature of water is from 20 °C to 25 °C. Remember that the formula of ΔT is:

`ΔT=|Final\text{ temperature-Initial temperature\mid}`

The change of temperature in celsius will be the same in kelvin. So our initial formula would be:

`\begin{gathered} Heat\text{ lost by iron=Heat gained by water} \\ c_(iron)\cdot m_(iron)\cdot\Delta T=c_(water)\cdot m_(water)\cdot\Delta T \\ 0.449\text{ }(J)/(g\cdot K)\cdot100g\cdot|25-150|K=c_(water)\cdot268.5g\cdot|25-20|K. \end{gathered}`

And finally, we solve for 'c' of water:

`\begin{gathered} 5612.5\text{ J=c}_(water)\cdot1342.5\text{ g}\cdot K, \\ c_(water)=\frac{5612.5\text{ J}}{1342.5\text{ g}\cdot K}, \\ c_(water)=4.18(J)/(g\cdot K). \end{gathered}`

The specific heat of water would be 4.18 J/gK.