If 200 grams of water is to be heated from 24°C to 100°C to make a cup of tea, how much heat must be added?

The specific heat of water is 4.18 #J##/##g*C#.

Answer 1

#q = "64 kJ"#

As you know, a substance's specific heat tells you how much heat is required to increase the temperature of #"1 g"# of that sample by #1^@"C"#.
In water's case, you know that its specific heat is equal to #4.18"J"/("g" ""^@"C")#. This tells you that in order to increase the temperature of #"1 g"# of water by #1^@"C"#, you need to supply #"4.18 J"# of heat.
How much heat would you need to increase the temperature of #"200 g"# of water by #1^@"C"#?
Well, if you need #"4.18 J"# per gram to increase its temperature by #1^@"C"#, it follows that you will need #200# times more heat to get this done.
Likewise, if you were to increase the temperature of #"1 g"# of water by #76^@"C"#, you'd need #76# times more heat than when increasing the temperature of #"1 g"# by #1^@"C"#.
Combine these two requirements and you get the total amount of heat required to increase the temperature of #"200 g"# of water by #76^@"C"#.

This can be expressed mathematically as the following equation.

#color(blue)(q = m * c * DeltaT)" "#, where
#q# - heat absorbed/lost #m# - the mass of the sample #c# - the specific heat of the substance #DeltaT# - the change in temperature, defined as final temperature minus initial temperature

Enter your values to obtain

#q = 200 color(red)(cancel(color(black)("g"))) * 4.18"J"/(color(red)(cancel(color(black)("g"))) color(red)(cancel(color(black)(""^@"C")))) * (100 - 24)color(red)(cancel(color(black)(""^@"C")))#
#q = "63536 J"#

I'll leave the response in kilojoules, rounded to two sig figs.

#q = color(green)("64 kJ")#
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Answer 2

The amount of heat required to raise the temperature of a substance can be calculated using the formula:

Q = mcΔT

Where:

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

For water, the specific heat capacity is approximately 4.18 J/g°C.

Given:

  • Mass of water (m) = 200 grams
  • Initial temperature (T_initial) = 24°C
  • Final temperature (T_final) = 100°C

Calculate the change in temperature (ΔT): ΔT = T_final - T_initial ΔT = 100°C - 24°C ΔT = 76°C

Now, plug in the values into the formula to calculate the heat energy (Q):

Q = (200 g) × (4.18 J/g°C) × (76°C) Q ≈ 63,536 J

Therefore, approximately 63,536 joules of heat energy must be added to raise the temperature of 200 grams of water from 24°C to 100°C.

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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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