Question

105 mL of H2O is initially at room temperature (22.0∘C). A chilled steel rod at 2.0∘C is placed in the water. If the final temperature of the system is 21.3 ∘C, what is the mass of the steel bar? Specific heat of water = 4.18 J/g⋅∘C Specific heat of steel = 0.452 J/g⋅∘C

Answer 1

Answer : The mass of the steel bar is, 35.2 grams.

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 steel = $0.452J/g^oC$

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

$m_1$ = mass of steel rod = ?

$m_2$ = mass of water  = $Density\times Volume=1.00g/mL\times 105mL=150g$

$T_f$ = final temperature of mixture = $21.3^oC$

$T_1$ = initial temperature of steel = $2.0^oC$

$T_2$ = initial temperature of water = $22.0^oC$

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

$m_1\times (0.452J/g^oC)\times (21.3-2.0)^oC=-(105g)\times 4.18J/g^oC\times (21.3-22.0)^oC$

$m_1=35.2g$

Therefore, the mass of the steel bar is, 35.2 grams.