Question

. 125g of water has an initial temperature of 25.6°C, and is heated by 50.0g of a metal

Answer 1

Answer : The specific heat of the metal is, $0.451J/g^oC$

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 = 50.0 g

$m_2$ = mass of water  = 125 g

$T_f$ = final temperature of mixture = $29.3^oC$

$T_1$ = initial temperature of metal = $115.0^oC$

$T_2$ = initial temperature of water = $25.6^oC$

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

$(50.0g)\times c_1\times (29.3-115.0)^oC=-[(125g)\times 4.18J/g^oC\times (29.3-25.6)^oC]$

$c_1=0.451J/g^oC$

Therefore, the specific heat of the metal is, $0.451J/g^oC$