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3 years ago

How much heat in kilojoules is required to melt an ice cube with a mass of 18.6 g at 0 °C? The Lf for water is 333 J/g.

1 answer:

3 0

Q = m.Lf.
Here, m = 18.6 g
Lf = 333 J/g

Substitute their values,
Q = 18.6 × 333 = 6193.8 J

We know, 1 KJ = 1000 J
So, 6193.8 J = 6.193 KJ

In short, Your Answer would be Option D

Hope this helps!

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Problem 6.056 Air enters a compressor operating at steady state at 15 lbf/in.2, 80°F and exits at 275°F. Stray heat transfer and

vova2212 [387]

To solve this process it is necessary to consider the concepts related to the relations between pressure and temperature in an adiabatic process.

By definition the relationship between pressure and temperature is given by

$(\frac{P_2}{P_1})=(\frac{T_2}{T_1})^{(\frac{\gamma}{\gamma-1})}$

Here

P = Pressure

T = Temperature

$\gamma =$ The ratio of specific heats. For air normally is 1.4.

Our values are given as,

$P_1 = 15lb/in^2\\T_1= 80\°F = 299.817K\\T_2 =400\°F = 408.15K$

Therefore replacing we have,

$(\frac{P_2}{P_1})=(\frac{T_2}{T_1})^{(\frac{\gamma}{\gamma-1})}$

$(\frac{P_2}{15})=(\frac{408.15}{299.817})^{(\frac{1.4}{1.4-1})}$

Solving for $P_2,$

$P_2 = 15*(\frac{408.15}{299.817})^{(\frac{1.4}{1.4-1})}$

$P_2 = 44.15Lbf/in^2$

Therefore the maximum theoretical pressure at the exit is $44.15Lbf/in^2$

5 0

3 years ago

What are the factors that affect the resistance of a wire?

777dan777 [17]

1) Length of the wire.

2) Thickness of the wire.

3) Temperature.

4) Type of metal.

Hope this helps!

-Payshence

6 0

3 years ago

ANSWER FOR BRAINLIEST

Nikitich [7]

A (b) would be 42 (c) from x to Y mark me as brainlist thanks

5 0

2 years ago

Carmen is helping load furniture and boxes onto a moving truck. She picks up boxes of her things, places them on a cart, and pus

AleksandrR [38]

Answer:

B because of the friction from the wheels

6 0

3 years ago

Read 2 more answers

The captain of a boat wants to travel directly across a river that flows due exst with a speed of 100 m/sHe starts from the sout

Salsk061 [2.6K]

Answer:

86.51° North of West or 273.49°

Explanation:

Let V' = velocity of boat relative to the earth, v = velocity of boat relative to water and V = velocity of water.

Now, by vector addition V' = V + v'.

Since v' = 6.10 m/s in the north direction, v' = (6.10 m/s)j and V = 100 m/s in the east direction, V = (100 m/s)i. So that

V' = V + v'

V' = (100 m/s)i + (6.10 m/s)j

So, we find the direction,Ф the boat must steer to from the components of V'.

So tanФ = 6.10 m/s ÷ 100 m/s

tanФ = 0.061

Ф = tan⁻¹(0.061) = 3.49°

So, the angle from the north is thus 90° - 3.49° = 86.51° North of West or 270° + 3.49° = 273.49°

7 0

3 years ago