Question

A 72.65-g bar of iron was heated to 99.7˚C and then dropped into 35.0 mL of water at 20.0˚C. The temperature of the water increased to a maximum of 33.2˚C. Calculate the specific heat of iron. Assume that the density of water is 1.00 g/mL.

Answer 1

Answer:

$C_{Fe}=0.096 \frac{cal}{g*C}$

Explanation:

The energy gained by the water:

$E=\Delta T*C*V* \rho$

$E=(33.2 - 20)C*\frac{1 cal}{g*C}*35 mL*1 g/mL$

$E=462 cal$

This energy is equal to the energy lost by the iron:

$E=\Delta T*C_{Fe}*m_{Fe}$

In equlibrium the T of the iron and the water are the same

$-462 cal=(33.2 - 99.7)C*C_{Fe}*72.65g$  

(the negative sign is because the iron loses the energy)

$C_{Fe}=0.096 \frac{cal}{g*C}$