Question

If a 95.0 gram sample of metal at 100.0 oC is placed in 50.0 g of water with an initial temperature of 22.5oC and the final temperature of the system is 48.5oC, what is the specific heat of the metal? Group of answer choices

Answer 1

Answer : The specific heat of the metal is, $1.11J/g^oC$

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 = 95.0g

$m_2$ = mass of water  = 50.0 g

$T_f$ = final temperature of mixture = $48.5^oC$

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

$T_2$ = initial temperature of water = $22.5^oC$

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

$(95.0g)\times c_1\times (48.5-100.0)^oC=-[(50.0g)\times 4.18J/g^oC\times (48.5-22.5)^oC]$

$c_1=1.11J/g^oC$

Therefore, the specific heat of the metal is, $1.11J/g^oC$