How many calories are absorbed when 500 grams of water (1 cal/g c) is heated from 50 to 100 degrees Celsius?

Answer 1

#"30,000 cal"#

You know that the specific heat of a substance tells you the amount of heat needed to increase the temperature of #"1 g"# of that substance by #1^@"C"#.

In your case, water is said to have a specific heat equal to

#c_"water" = "1 cal g"^(-1)""^@"C"^(-1)#
This means that in order to increase the temperature of #"1 g"# of water by #1^@"C"#, you need to provide it with #"1 cal"# of heat.
So, how much heat would be needed to increase the temperature of #"500 g"# of water by #1^@"C"# ?
#500 color(red)(cancel(color(black)("g"))) * overbrace("1 cal"/(1color(red)(cancel(color(black)("g")))))^(color(blue)("for 1"^@"C")) = "500 cal"#
This means that in order to increase the temperature of #"500 g"# of water by
#100^@"C" - 50^@"C" = 50^@"C"#

you need to provide it with

#50 color(red)(cancel(color(black)(""^@"C"))) * overbrace("500 cal"/(1color(red)(cancel(color(black)(""^@"C")))))^(color(blue)("for 500 g")) = "25,000 cal" = color(darkgreen)(ul(color(black)("30,000 cal")))#

The answer must be rounded to one significant figure, the number of sig figs you have for your values.

Keep in mind that this much heat is needed in order to get #"500 g"# of water from liquid at #50^@"C"# to liquid at #100^@"C"#.
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Answer 2

To calculate the calories absorbed when heating water from 50 to 100 degrees Celsius, you can use the formula:

Q = mcΔT

Where: Q = heat energy absorbed (in calories) m = mass of the water (in grams) c = specific heat capacity of water (1 cal/g°C) ΔT = change in temperature (final temperature - initial temperature)

Given: m = 500 grams c = 1 cal/g°C ΔT = (100°C - 50°C) = 50°C

Now, substitute the values into the formula:

Q = (500 g) * (1 cal/g°C) * (50°C) = 25,000 calories

So, 25,000 calories are absorbed when 500 grams of water is heated from 50 to 100 degrees Celsius.

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Answer 3

To calculate the calories absorbed when heating 500 grams of water from 50 to 100 degrees Celsius, you can use the formula:

[ Q = mcΔT ]

Where:

  • ( Q ) is the heat absorbed (in calories)
  • ( m ) is the mass of the substance (in grams)
  • ( c ) is the specific heat capacity of the substance (in cal/g°C)
  • ( ΔT ) is the change in temperature (in °C)

Substituting the given values:

( m = 500 ) grams ( c = 1 ) cal/g°C ( ΔT = (100 - 50) = 50 ) °C

[ Q = (500 , \text{g}) \times (1 , \text{cal/g°C}) \times (50 , \text{°C}) ]

[ Q = 500 \times 1 \times 50 ]

[ Q = 25000 ]

So, 25,000 calories are absorbed when 500 grams of water is heated from 50 to 100 degrees Celsius.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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