Chemistry, 20.03.2020 05:58, soonerlady19

A volume of 120 mL of H2O is initially at room temperature (22.00 ∘C ). A chilled steel rod at 2.00 ∘C ∘C is placed in the water. If the final temperature of the system is 21.50 ∘C ∘C , what is the mass of the steel bar? Use the following values: specific heat of water = 4.18 J/(g⋅∘C)J/(g⋅∘C) specific heat of steel = 0.452 J/(g⋅∘C)

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answered: alejandrosaaved1

The mass of the steel bar is "28.5 g"

According to the question,

Mass,

m = 120 mL

Specific heat of steel,

c = 0.452 J/(g⋅°C)

Specific heat of water,

c = 4.18 J/(g⋅°C)

Temperature,

2°C22°C21.5°C

As we know,

→ Heat lost by water = Heat gained by steel

then,

→ $mc \Delta T = mc \Delta T$

By substituting the values, we get

→ $120\times 4.18\times (22-21.5) = m\times 0.452\times (21.5-2)$

→ $m = \frac{120\times 4.18\times 0.5}{0.452\times 19.5}$

→ $= \frac{250.8}{8.814}$

→ $= 28.5 \ g$

Thus the above answer is appropriate.

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answered: trevorpeterson20

$\large \boxed{\text{28.5 g}}$

Explanation:

There are two heat flows in this process and, since energy (heat) can neither be destroyed nor created, the energy change for the system must equal zero.

Data:

For Fe, m₁ = ?; C₁ = 0.452 J°C⁻¹g⁻¹; Ti = 2.00 °C; T_f = 21.50 °C

For H₂O, m₂ = 120 g; C₂ = 4.18 J°C⁻¹g⁻¹; Ti = 22.00 °C; T_f = 21.50 °C

Calculations:

1. Temperature changes

ΔT₁ = T_f - Ti = 21.50 °C - 2.00 °C = 19.50 °C

ΔT₂ = T_f - Ti = 21.50 °C - 22.00 °C = -0.50 °C

2. Mass of steel rod

$\begin{array}{ccl}\text{Heat gained by steel rod + heat lost by water} & = & 0\\m_{1}C_{1} \Delta T_{1} + m_{2}C_{2} \Delta T_{2}& = & 0\\m_{1} \times 0.452 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$}\times 19.50 \, ^{\circ}\text{C} + \text{120 g} \times 4.18\text{ J$^{\circ}$C$^{-1}$g$^{-1}$} \times (-0.50)\, ^{\circ}\text{C}& = & 0\\8.814m_{1}\text{ g}^{-1} - 250.8 & = &0\\\end{array}\\$

$\begin{array}{ccl}8.814m_{1}\text{ g}^{-1} & = &250.8\\m_{1} & = & \dfrac{250.8}{\text{8.814 g}^{-1}}\\\\& = & \textbf{28.5 g}\\\end{array}\\\text{The mass of the steel rod is $\large \boxed{\textbf{28.5 g}}$}$

answered: Guest

It has to be b i hope this

answered: Guest

hydrocarbons are organic compounds that contain only hydrogens and carbons. they can be saturated (those which contain only carbon-carbon single bonds) or unsaturated (those which contain carbon-carbon double or triple bonds) in nature. hydrocarbons can undergo several reactions like substitution, elimination etc.

on substituting one or more hydrogen atoms in hydrocarbon it results in the formation of haloalkane. as the halogen atoms are large compared to the carbon and hydrogen atoms, due to which the molecular weight increases and the bond become polar because of presence electronegative halogen atom and results in the increase in boiling point of the haloalkane.

thus, the correct option is (b) that is the boiling point of the new compound increases on substituting a hydrogen atom with a halogen in a hydrocarbon.

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A volume of 120 mL of H2O is initially at room temperature (22.00 ∘C ). A chilled steel rod at 2.00...