Question

A hot 105.8 g lump of an unknown substance initially at 179.7 °C is placed in 35.0 mL of water initially at 25.0 °C and the system is allowed to reach thermal equilibrium. The final temperature of the system is 77.4 °C. Substance Specific heat (J/(g·°C)) aluminum 0.897 graphite 0.709 rhodium 0.243 titanium 0.523 tungsten 0.132 zinc 0.388 water 4.184 Using this information and the specific heat values for several metals in the table, identify the unknown substance. Assume no heat is lost to the surroundings. tungsten aluminum titanium zinc graphite rhodium

Answer 1

m(water) = 35.0 g

T(water) = 25.0 oC

C(water) = 4.184 J/goC

m(lump) = 105.8 g

T(lump) = 179.7 oC

C(lump) = to be calculated

We will be using heat conservation equation

use:

heat lost by lump = heat gained by water

m(lump)*C(lump)*(T(lump)-T) = m(water)*C(water)*(T-T(water))

105.8*C(lump)*(179.7-77.4) = 35.0*4.184*(77.4-25.0)

10823.34*C(lump) = 7673.456

C(lump)= 0.709 J/goC

This is specific heat capacity of graphite

Answer: graphite