Question

You placed 43.1 g of an unknown metal at 100 °C into a coffee cup calorimeter that contained 50.0 g of water that was initially at 22.0 °C. The equilibrium temperature of mixing (T0) was determined to be 23.2 °C. The calorimeter constant was known to be 51.5 J/°C. Specific HeatH2O = 4.184 J/g·°Ca. What is the total amount of heat (J) lost by the metal? NG 1.5b. What was the specific heat (J/g·°C) of the metal? NG 1.5

Answer 1

Answer :

(a) The heat released by the metal is -312.48 J

(b) The specific heat of the metal is $0.0944J/g^oC$

Explanation :

For part A :

Heat released by the metal = Heat absorbed by the calorimeter + Heat absorbed by the water

$q=[q_1+q_2]$

$q=[c_1\times \Delta T+m_2\times c_2\times \Delta T]$

where,

q = heat released by the metal

$q_1$ = heat absorbed by the calorimeter

$q_2$ = heat absorbed by the water

$c_1$ = specific heat of calorimeter = $51.5J/^oC$

$c_2$ = specific heat of water = $4.184J/g^oC$

$m_2$ = mass of water = 50.0 g

$\Delta T$ = change in temperature = $(T_{final}-T_{initial})=23.2-22.0=1.2^oC$

Now put all the given values in the above formula, we get:

$q=[(51.5J/^oC\times 1.2^oC)+(50.0g\times 4.184J/g^oC\times 1.2^oC)]$

$q=312.48J$

Thus, the heat released by the metal is -312.48 J

For part B :

$q=m\times c\times \Delta T$

q = heat released by the metal = -312.48 J

m = mass of metal = 43.1 g

c = specific heat of metal = ?

$\Delta T$ = change in temperature = $(T_{final}-T_{initial})=23.2-100=76.8^oC$

Now put all the given values in the above formula, we get:

$-312.48J=43.1g\times c\times 76.8^oC$

$c=0.0944J/g^oC$

Thus, the specific heat of the metal is $0.0944J/g^oC$