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* c_1* (T_f-T_1)=-m_2* c_2* (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* Volume=1.00g/mL* 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* (0.452J/g^oC)* (21.3-2.0)^oC=-(105g)* 4.18J/g^oC* (21.3-22.0)^oC`

`m_1=35.2g`

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