Question

If 75 g of metal at 100.0∘C is transferred into 80.0 mL of water at 20.0∘C initially. The water is then raised to 25∘C. What is the specific heat of the metal?

Answer 1

Answer : The specific heat of the metal is, $0.297J/g^oC$

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

$q_1=-q_2$

$m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)$

where,

$c_1$ = specific heat of metal = ?

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

$m_1$ = mass of metal = 75 g

$m_2$ = mass of water  = $Density\times Volume=1.00g/mL\times 80.0mL=80.0g$

$T_f$ = final temperature of mixture = $25^oC$

$T_1$ = initial temperature of metal = $100.0.0^oC$

$T_2$ = initial temperature of water = $20.0^oC$

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

$(75g)\times c_1\times (25-100.0)^oC=-[(80.0g)\times 4.18J/g^oC\times (25-20.0)^oC]$

$c_1=0.297J/g^oC$

Therefore, the specific heat of the metal is, $0.297J/g^oC$