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3 years ago

What is the amount of heat, in Sl units, necessary to melt 1 lb of ice?

Physics

1 answer:

likoan [24]3 years ago

3 0

Answer:

Heat required to melt 1 lb of ice is 151.469 KJ

Explanation:

We have given mass of ice = 1 lb

We know that 1 lb = 0.4535 kg

Latent heat of fusion for ice =334 KJ/kg

Amount if heat for fusion of ice is given by

$Q=mL$ , here m is mass of ice and L is latent heat of fusion

So heat $Q=mL=0.4535\times 334=151.469kj$

So heat required to melt 1 lb of ice is equal to 151.469 KJ

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Pre-Test Active

vlada-n [284]

n is the answer I guess

not so sure

7 0

2 years ago

How can icalculate 3 seconds and 5 meters into a speed?

topjm [15]

To find speed you have to divide distance by time. In this case:

5 meters➗3 seconds = about 1.66666666 and so on m/s.

You could round to 1.67 or 1.7 if you'd like.

8 0

3 years ago

The length of aluminum rods produced by a company are approximated by a Gaussian distribution with a mean of 10 cm and a standar

ExtremeBDS [4]

Given Information:

Mean length of aluminum rods = μ = 10 cm

Standard deviation of length of aluminum rods = σ = 0.02 cm

Required Information:

a) P(9.98 < X < 10.02) = ?

b) P(9.90 < X < 10.1) = ?

Answer:

a) P(9.98 < X < 10.02) = 68.27%

b) P(9.90 < X < 10.1) = 100%

Explanation:

What is Normal Distribution?

Normal Distribution or also known as Gaussian Distribution, is a continuous probability distribution and is symmetrical around the mean. The shape of this distribution is like a bell curve and most of the data is clustered around the mean. The area under this bell shaped curve represents the probability

a) We want to find out the probability that the length of aluminum rods is between 9.98 and 10.02 cm.

$P(9.98 < X < 10.02) = P( \frac{x - \mu}{\sigma} < Z < \frac{x - \mu}{\sigma} )\\\\P(9.98 < X < 10.02) = P( \frac{9.98- 10}{0.02} < Z < \frac{10.02 - 10}{0.02} )\\\\P(9.98 < X < 10.02) = P( \frac{-0.02}{0.02} < Z < \frac{0.02}{0.02} )\\\\P(9.98 < X < 10.02) = P( -1 < Z < 1 )\\\\$

The z-score corresponding to -1 is 0.15866 and 1 is 0.84134

$P(9.98 < X < 10.02) = P( Z < 1 ) - P( Z < -1 ) \\\\P(9.98 < X < 10.02) = 0.84134 - 0.15866 \\\\P(9.98 < X < 10.02) = 0.6827\\\\P(9.98 < X < 10.02) = 68.27 \%$

Therefore, the probability that the length of aluminum rods is between 9.98 and 10.02 cm is 68.27%

b) We want to find out the probability that the length of aluminum rods is between 9.90 and 10.1 cm.

$P(9.90 < X < 10.1) = P( \frac{x - \mu}{\sigma} < Z < \frac{x - \mu}{\sigma} )\\\\P(9.90 < X < 10.1) = P( \frac{9.90- 10}{0.02} < Z < \frac{10.1 - 10}{0.02} )\\\\P(9.90 < X < 10.1) = P( \frac{-0.1}{0.02} < Z < \frac{0.1}{0.02} )\\\\P(9.90 < X < 10.1) = P( -5 < Z < 5 )\\\\$

The z-score corresponding to -5 is 0 and 5 is 1

$P(9.90 < X < 10.1) = P( Z < 5 ) - P( Z < -5 ) \\\\P(9.90 < X < 10.1) = 1 - 0 \\\\P(9.90 < X < 10.1) = 1\\\\P(9.90 < X < 10.1) = 100 \%$

Therefore, the probability that the length of aluminum rods is between 9.90 and 10.1 cm is 100%

How to use z-table?

Step 1:

In the z-table, find the two-digit number on the left side corresponding to your z-score. (e.g 1.0, 2.2, 0.5 etc.)

Step 2:

Then look up at the top of z-table to find the remaining decimal point in the range of 0.00 to 0.09. (e.g. if you are looking for 1.00 then go for 0.00 column)

Step 3:

Finally, find the corresponding probability from the z-table at the intersection of step 1 and step 2.

8 0

3 years ago

Given that no sunlight can penetrate Europa’s ice shell, what would be the type of energy that could make some form of europan l

Anna71 [15]

Answer:

life in europa is possible due to chemical energy, surface and ocean intracts with one another and give off chemical energy.

Explanation:

the ocean in europa is in contact with the underlying rocky mantle and at high temperatures, its rocks and water interact. oxidization and reduction in which molecules give off and accept electrons occur in it and in deep ocean, when chemically reduction vent fluids interacts with oxygen containing seawater to produce energy that can support micro-organisms and animals.

5 0

3 years ago

(11%) Problem 5: A uniform stationary ladder of length L = 2.7 m and mass M = 11 kg leans against a smooth vertical wall, while

jeka94

Answer:

970.2 N

Explanation:

We are given that

Length of ladder=2.7 m

Mass,M=11 kg

Coefficient of friction= $\mu=0.45$