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

Answer:

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

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:

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

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

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

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