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Ksju [112]
3 years ago
15

Here, we want to become proficient at changing units so that we can perform calculations as needed. The basic heat transfer equa

tion is q = m*Cp*T, where q is the amount of heat added or subtracted, Cp is the constant pressure heat capacity, and T is the change in temperature (T = T2 – T1). For example, we need to heat 100 lbs of water from 25 °C to 75 °C. How much energy must we add to the water if the constant pressure heat capacity of water is 4.18 J/g K? (Give answer in both kJ and BTU)
Engineering
1 answer:
netineya [11]3 years ago
7 0

Answer:

9500 kJ; 9000 Btu

Explanation:

Data:

m = 100 lb

T₁ = 25 °C

T₂ = 75 °C

Calculations:

1. Energy in kilojoules

ΔT = 75 °C - 25 °C = 50 °C  = 50 K

m = \text{100 lb} \times \dfrac{\text{1 kg}}{\text{2.205 lb}} \times \dfrac{\text{1000 g}}{\text{1 kg}}= 4.54 \times 10^{4}\text{ g}\\\\\begin{array}{rcl}q & = & mC_{\text{p}}\Delta T\\& = & 4.54 \times 10^{4}\text{ g} \times 4.18 \text{ J$\cdot$K$^{-1}$g$^{-1}$} \times 50 \text{ K}\\ & = & 9.5 \times 10^{6}\text{ J}\\ & = & \textbf{9500 kJ}\\\end{array}

2. Energy in British thermal units

\text{Energy} = \text{9500 kJ} \times \dfrac{\text{1 Btu}}{\text{1.055 kJ}} = \text{9000 Btu}

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telo118 [61]

Answer:

a) the inductance of the coil is 6 mH

b) the emf generated in the coil is 18 mV  

Explanation:

Given the data in the question;

N = 570 turns

diameter of tube d = 8.10 cm = 0.081 m

length of the wire-wrapped portion l =  35.0 cm = 0.35 m

a) the inductance of the coil (in mH)

inductance of solenoid

L = N²μA / l

A = πd²/4  

so

L = N²μ(πd²/4) / l

L = N²μ(πd²) / 4l

we know that μ = 4π × 10⁻⁷ TmA⁻¹

we substitute

L = [(570)² × 4π × 10⁻⁷× ( π × (0.081)² )] / 4(0.35)

L =  0.00841549 / 1.4

L = 6 × 10⁻³ H    

L = 6 × 10⁻³ × 1000 mH

L = 6 mH

Therefore, the inductance of the coil is 6 mH

b)

Emf ( ∈ ) = L di/dt

given that; di/dt = 3.00 A/sec

{∴ di = 3 - 0 = 3 and dt = 1 sec}

Emf ( ∈ ) = L di/dt

we substitute

⇒ 6 × 10⁻³ ( 3/1 )

= 18 × 10⁻³ V

= 18 × 10⁻³ × 1000

= 18 mV  

Therefore, the emf generated in the coil is 18 mV  

7 0
2 years ago
Water flows through a pipe at an average temperature of T[infinity] = 70°C. The inner and outer radii of the pipe are r1 = 6 cm
Paul [167]

Answer:

The differential equation and the boundary conditions are;

A) -kdT(r1)/dr = h[T∞ - T(r1)]

B) -kdT(r2)/dr = q'_s = 734.56 W/m²

Explanation:

We are given;

T∞ = 70°C.

Inner radii pipe; r1 = 6cm = 0.06 m

Outer radii of pipe;r2 = 6.5cm=0.065 m

Electrical heat power; Q'_s = 300 W

Since power is 300 W per metre length, then; L = 1 m

Now, to the heat flux at the surface of the wire is given by the formula;

q'_s = Q'_s/A

Where A is area = 2πrL

We'll use r2 = 0.065 m

A = 2π(0.065) × 1 = 0.13π

Thus;

q'_s = 300/0.13π

q'_s = 734.56 W/m²

The differential equation and the boundary conditions are;

A) -kdT(r1)/dr = h[T∞ - T(r1)]

B) -kdT(r2)/dr = q'_s = 734.56 W/m²

6 0
2 years ago
Assume you have four fins, each with a mass of 8.0 grams. What is the total weight of these four fins? (Hint: watch your units!)
soldier1979 [14.2K]
The answer is 32.0 grams
7 0
2 years ago
A simple formula to estimate the upward velocity of a rocket (neglecting the aerodynamic drag) is:
Bingel [31]

Answer:

Test code:

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Explanation:

See attached image

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olga55 [171]

Explanation:

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8 0
2 years ago
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