Question

A 11.6 g piece of metal is heated to 98°C and dropped into a calorimeter containing 50.0 g of water (specific heat capacity of water is 4.18 J/g°C) initially at 20.5°C. The empty calorimeter has a heat capacity of 125 J/K. The final temperature of the water is 28.2°C. Ignoring significant figures, calculate the specific heat of the metal.

Answer 1

Answer:2

Explanation:

Answer 2

Answer : The specific heat of the metal is, $3.18J/g^oC$

Explanation :

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

$q=-[q_1+q_2]$

$m\times c\times (T_f-T_1)=-[c_1\times (T_f-T_2)+m_2\times c_2\times (T_f-T_2)]$

where,

q = heat released by the reaction

$q_1$ = heat absorbed by the calorimeter

$q_2$ = heat absorbed by the water

c = specific heat of metal = ?

$c_1$ = specific heat of calorimeter = $125J/^oC$

$c_2$ = specific heat of water = $4.18J/g^oC$

$m_2$ = mass of water = 50.0 g

m = mass of metal = 11.6 g

$T_f$ = final temperature = $28.2^oC$

$T_1$ = temperature of metal = $98^oC$

$T_2$ = temperature of water = $20.5^oC$

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

$11.6g\times c\times (28.2-98)^oC=-[125J/^oC\times (28.2-20.5)^oC+50.0g\times 4.18J/g^oC\times (28.2-20.5)^oC]$

$c=3.18J/g^oC$

Thus, the specific heat of the metal is, $3.18J/g^oC$