Question

n unknown metal is either aluminum, iron or lead. If 150. g of this metal at 150.0 °C was placed in a calorimeter that contains 200. g of water at 25.0 °C and the final temperature was found to be 34.3 °C after thermal equilibrium was achieved, assuming heat was only transferred between water and metal, what is the identity of this metal? Some specific heat values of metals and water given below may be useful. Specific heats, J/(g•°C): Fe (0.449) Pb (0.128) Al (0.903) H2O (4.184)

Answer 1

Answer : The metal used was iron (the specific heat capacity is $0.449J/g^oC$).

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 unknown metal = ?

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

$m_1$ = mass of unknown metal = 150 g

$m_2$ = mass of water = 200 g

$T_f$ = final temperature of water = $34.3^oC$

$T_1$ = initial temperature of unknown metal = $150.0^oC$

$T_2$ = initial temperature of water = $25.0^oC$

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

$150g\times c_1\times (34.3-150.0)^oC=-200g\times 4.184J/g^oC\times (34.3-25.0)^oC$

$c_1=0.449J/g^oC$

Form the value of specific heat of unknown metal, we conclude that the metal used in this was iron (Fe).

Therefore, the metal used was iron (the specific heat capacity is $0.449J/g^oC$).