Question

Please help!

When a 200.0 g block of iron is placed into 1500.0 g of water, the temperature of the iron increases by 50.0 °C. The specific heat of iron is 0.449 J/g^ C . What is the temperature change of the water? (0.715 C)
Thank you!

Answer 1

`\Delta T_(w)=-0.715\°C`

Step-by-step explanation:

Hello there!

In this case, according to the given information, it is possible to realize that the heat relationship between iron and water is given by:

`-Q_(Fe)=Q_(w)`

Which in terms of mass, specific heat and temperature is:

`-m_(Fe)C_(Fe)\Delta T_(Fe)=m_(w)C_(w)\Delta T_(w)`

Thus, by solving for the change in the temperature of water we will obtain:

`\Delta T_(w)=(-m_(Fe)C_(Fe)\Delta T_(Fe))/(m_(w)C_(w)) \\\\\Delta T_(w)=(-200.0g*0.449(J)/(g\°C) 50.0\°C)/(1500.0g4.184(J)/(g\°C)) \\\\\Delta T_(w)=-0.715\°C`

Which is negative because iron increases its temperature and therefore water decreases it.

Regards!