Question

A volume of 105 mL of H2O is initially at room temperature (22.00 ∘C). A chilled steel rod at 2.00 ∘C is placed in the water. If the final temperature of the system is 21.50 ∘C , what is the mass of the steel bar? Use the following values: specific heat of water = 4.18 J/(g⋅∘C) specific heat of steel = 0.452 J/(g⋅∘C) Express your answer to three significant figures and include the appropriate units.

Answer 1

25.0 grams is the mass of the steel bar.

Step-by-step explanation:

Heat gained by steel bar will be equal to heat lost by the water

`Q_1=-Q_2`

Mass of steel=
`m_1`

Specific heat capacity of steel =
`c_1=0.452 J/g^oC`

Initial temperature of the steel =
`T_1=2.00^oC`

Final temperature of the steel =
`T_2=T=21.50^oC`

`Q_1=m_1c_1* (T-T_1)`

Mass of water=
`m_2= 105 g`

Specific heat capacity of water=
`c_2=4.18 J/g^oC`

Initial temperature of the water =
`T_3=22.00^oC`

Final temperature of water =
`T_2=T=21.50^oC`

`Q_2=m_2c_2* (T-T_3)-Q_1=Q_2(m_1c_1* (T-T_1))=-(m_2c_2* (T-T_3))`

On substituting all values:

`(m_1* 0.452 J/g^oC* (21.50^o-2.00^oC))=-(105 g* 4.18 J/g^oC* (21.50^o-22.00^o))\\\\m_1*8.7914=241.395\\\\m_1=(219.45)/(8.7914) \\\\m_1=24.9\\\\ \approx25 \texttt {grams}`

25.0 grams is the mass of the steel bar.