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faltersainse [42]
2 years ago
8

The Chemistry Cafe was out of bread. The cook went next door to the bakery and bought a loaf of bread which has 33 slices. Then,

she sells 12 sandwiches (two pieces of bread per sandwich). How much bread did she have left? Explain your answer.​
Chemistry
2 answers:
maria [59]2 years ago
4 0

In the question, we are told that there are;

  • A loaf containing 33 slices
  • A loaf containing 33 slices A package of cheese containing 15 slices

We also know that he is making a sandwich that has 2 pieces of both cheese and bread.

Hence;

Total number of bread and cheese = 33 + 15.

Each loaf should have two pieces of each bread and the cheeses make a total of four pieces.

Therefore he can make = 33 + 15/4 = 12 sandwiches.

scoundrel [369]2 years ago
4 0
The cook uses 2 slices of bread per sandwich. Which is equivalent to (the slices of bread) x (the total number of sandwiches) = total number of bread slices used. In this case 2 x 12 =24 slices used. Since the loaf of bread had 33, subtract the slices of bread used, 24. This results in 33-24= 9 slices of bread remaining.
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How can one kg of iron melt more ice than 1 kg lead at 100 °C
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Due to the specific heat capacity of iron, 0.444 J/(g·°C), is more than the specific heat capacity for lead, 0.160 J/(g·°C)

Explanation:

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The metals provided to melt the ice and their temperature includes;

One kg (1000 g) of iron;

Specific heat capacity = 0.444 J/(g·°C)

Temperature = 100°C

1 kg (1000 g) of lead

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Temperature = 100°C

Therefore, the heat provided to the ice of mass m, and latent heat of 334 J/g at 0°C by the metals are as follows;

For iron, we have;

ΔQ = Mass × Specific heat capacity × Temperature change

ΔQ_{iron} = Heat obtained from the iron by the ice

ΔQ_{iron} = 0.444 m × 1000 × (100 - 0) = 44400 J

Heat absorbed by the ice for melting, H_l = Heat obtained from the iron

∴ Heat absorbed by the ice for melting, H_l = Mass of ice × Latent heat of ice

H_l = Mass of ice × 334 J/g = 44400 J

Mass of ice melted by the iron = 44400 J/334 (J/g) ≈ 132.9 g

Mass of ice melted by the iron ≈ 132.9 g

For lead, we have;

ΔQ = Mass × Specific heat capacity × Temperature change

ΔQ_{lead} = Heat obtained from the iron by the ice

ΔQ_{lead} = 0.160 m × 1000 × (100 - 0) = 16000 J

Heat absorbed by the ice for melting, H_l = Heat obtained from the iron

∴ Heat absorbed by the ice for melting, H_l = Mass of ice × Latent heat of ice

H_l = Mass of ice × 334 J/g = 16000 J

Mass of ice melted by the lead = 16000 J/334 (J/g) ≈ 47.9 g

Mass of ice melted by the lead ≈ 47.9 g

Therefore, mass of  ice melted by the iron, approximately 132.9 g, is more than mass of ice melted by the lead, approximately 47.9 g.

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