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3 years ago

43.278 kg - 28.1 g use significant figures rule

1 answer:

3 0

You cannot directly subtract the given values because their units are different. To be consistent, let's convert kilograms to grams with the conversion that 1,000 g = 1 kg.

43.278 kg(1,000 g/1 kg) - 28.1 g = 43249.9 grams

In subtraction, you will base the number of significant figures to the least of the given data. Since 28.1 has 3 significant figures, the answer should also contain 3 significant figures. Thus, it is much more convenient to report it in terms of kilograms.

43249.9 grams * 1 kg/ 1,000 g = 43.2 kg

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sweet-ann [11.9K]

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B) 80 cal

C) 540 cal

Explanation:

The amount of heat energy needed to raise the temperature of a certain mass of a substance is given by

$Q=mC\Delta T$

where

m is the mass of the substance

C is the specific heat capacity

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In this problem:

m = 1 g is the mass of water

$C=1 cal/g^{\circ}C$ is the specific heat capacity of water

$\Delta T=1^{\circ}C$ is the change in temperature

So, the heat needed is

$Q=(1)(1)(1)=1 cal$

For a solid substance at its melting point, the amount of heat needed to melt completely the substance is given by

$Q=m\lambda_f$

where

m is the mass of the substance

$\lambda_f$ is the specific latent heat of fusion of the substance

In this problem:

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- Specific latent heat of fusion of ice: $\lambda_f=80 cal/g$

So, the heat needed is

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For a liquid substance at its boiling point, the amount of heat needed to boil completely the substance is given by

$Q=m\lambda_v$

where

m is the mass of the substance

$\lambda_v$ is the specific latent heat of vaporization of the substance

In this problem:

- The water is already at boiling point, 100 °C

- Mass of the water: $m=1g$

- Specific latent heat of vaporization of water: $\lambda_v=540 cal/g$

So, the heat needed is

$Q=(1)(540)=540 cal$

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