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3 years ago

A 12 gram piece of metal is heated to 300 °C from 100 °C with 1120 Joules of energy. What is the specific heat of the metal?

Chemistry

1 answer:

Sergeeva-Olga [200]3 years ago

7 0

Answer:

The specific heat for the metal is 0.466 J/g°C.

Explanation:

Given,

Q = 1120 Joules

mass = 12 grams

T₁ = 100°C

T₂ = 300°C

The specific heat for the metal can be calculated by using the formula

Q = (mass) (ΔT) (Cp)

ΔT = T₂ - T₁ = 300°C - 100°C = 200°C

Substituting values,

1120 = (12)(200)(Cp)

Cp = 0.466 J/g°C.

Therefore, specific heat of the metal is 0.466 J/g°C.

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3 years ago

2- A sheet of steel 3.0-mm thick has nitrogen atmospheres on both sides at 900Â°C and is permitted to achieve a steady-state dif

Bezzdna [24]

Answer:

2.78x10⁻³ m

Explanation:

Nitrogen diffusion in the steel can be expressed using the following equation:

$J = - D*\frac{\Delta C}{\Delta x} = - D*\frac{C_{1} - C_{2}}{x_{1} - x_{2}}$

<u>Where:</u>

J: is the diffusion flux = 1.0x10⁻⁷ kgm⁻²s⁻¹

D: is the diffusion coefficient = 1.85x10⁻¹⁰ m²/s

C₁: is the concentration of nitrogen in the steel at the surface = 2 kg/m³

C₂: is the concentration of nitrogen at the distance x₂ = 0.5 kg/m³

x₁: is the distance at the surface = 0

x₂: is the distance to find

Hence, we have:

$J = - D*\frac{C_{1} - C_{2}}{- x_{2}}$ (1)

Solving equation (1) for x₂:

$x_{2} = D \frac{C_{1} - C_{2}}{J} = 1.85 \cdot 10^{-10} m^{2}/s \frac{2 kg/m^{3} - 0.5 kg/m^{3}}{1.0 \cdot 10^{-7} kgm^{-2}s^{-1}} = 2.78 \cdot 10^{-3} m$

Therefore, the nitrogen will enter 2.78x10⁻³ m into the sheet.

I hope it helps you!

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3 years ago