Chemistry College

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.

Answers

Answer 1

Answer:

25.0 grams is the mass of the steel bar.

Explanation:

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

[tex]Q_1=-Q_2[/tex]

Mass of steel= [tex]m_1[/tex]

Specific heat capacity of steel = [tex]c_1=0.452 J/g^oC[/tex]

Initial temperature of the steel = [tex]T_1=2.00^oC[/tex]

Final temperature of the steel = [tex]T_2=T=21.50^oC[/tex]

[tex]Q_1=m_1c_1\times (T-T_1)[/tex]

Mass of water= [tex]m_2= 105 g[/tex]

Specific heat capacity of water=[tex]c_2=4.18 J/g^oC[/tex]

Initial temperature of the water = [tex]T_3=22.00^oC[/tex]

Final temperature of water = [tex]T_2=T=21.50^oC[/tex]

[tex]Q_2=m_2c_2\times (T-T_3)-Q_1=Q_2(m_1c_1\times (T-T_1))=-(m_2c_2\times (T-T_3))[/tex]

On substituting all values:

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

25.0 grams is the mass of the steel bar.
Answer 2

Answer:

[tex]m_{steel}=24.9g[/tex]

Explanation:

Hello,

In this case, since the water is initially hot, the released heat by it is gained by the steel rod since it is initially cold which in energetic terms is illustrated by:

[tex]\Delta H_{water}=-\Delta H_{steel}[/tex]

That in terms of mass, specific heat and temperature change is:

[tex]m_{water}Cp_{water}(T_f-T_{water})=-m_{steel}Cp_{steel}(T_f-T_{steel})[/tex]

Thus, we simply solve for the mass of the steel rod:

[tex]m_{steel}=\frac{m_{water}Cp_{water}(T_f-T_{water})}{-Cp_{steel}(T_f-T_{steel})} \\\\m_{steel}=\frac{105mL*\frac{1g}{1mL}*4.18\frac{J}{g\°C}*(21.50-22.00)\°C}{-0.452\frac{J}{g\°C}*(21.50-2.00)\°C} \\\\m_{steel}=24.9g[/tex]

Best regards.

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