Question

A hot lump of 39.9 g of iron at an initial temperature of 78.1 °C is placed in 50.0 mL H 2 O initially at 25.0 °C and allowed to reach thermal equilibrium. What is the final temperature of the iron and water, given that the specific heat of iron is 0.449 J/(g⋅°C)? Assume no heat is lost to surroundings.

Answer 1

Answer : The final temperature of the mixture is $29.6^oC$

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 iron = $0.499J/g^oC$

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

$m_1$ = mass of iron = 39.9 g

$m_2$ = mass of water  = $Density\times Volume=1g/mL\times 50.0mL=50.0g$

$T_f$ = final temperature of mixture = ?

$T_1$ = initial temperature of iron = $78.1^oC$

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

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

$(39.9g)\times (0.499J/g^oC)\times (T_f-78.1)^oC=-(50.0g)\times 4.18J/g^oC\times (T_f-25.0)^oC$

$T_f=29.6^oC$

Therefore, the final temperature of the mixture is $29.6^oC$