Question

42.5 grams of an unknown substance is heated to 105.0 degrees Celsius and then placed into a calorimeter containing 110.0 grams of water at 24.2 degrees Celsius. If the final temperature reached in the calorimeter is 32.4 degrees Celsius, what is the specific heat of the unknown substance?

Show or explain the work needed to solve this problem, and remember that the specific heat capacity of water is 4.18 J/(°C x g).

Answer 1

q = mcΔT 
 = 110.0g * 4.18J/(°C x g) * 8.2°C
 = 3770.36 J

3770.36 J = 42.5 g * c * 72.6°C

c = 1.222 J/(°C x g)

Answer 2

Answer:The specific heat of the unknown substance is 1.22 J/ °Cg.

Explanation:

Heat absorbed by the water .

Mass of the water = 110.0 g

Change in temperature of water = $\Delta T=32.4^oC-24.2^oC=8.2^oC$

$Q=mc\Delta T=110.0g\times 4.18 J/^oCg\times 8.2^oC=3,770.36 J$

Heat lost by substance(Q') = Heat gained by the water(Q)

-Q' = Q

Change in temperature of the substance =

$\Delta T'=(32.4^oC)-105.0^oC=-72.6 ^oC$

Mass of the substance = m'=42.5 g

Specific heat of substance = c'

$-(m'\times c'\times \Delta T')=3,770.36 J$

$c'=\frac{3,770.36 J}{42.5 g\times 72.6^oC}=1.22 J/^oCg$

The specific heat of the unknown substance is 1.22 J/ °Cg.