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# .
The specific heat of water is 4.18
This can be expressed mathematically as the following equation.
Enter your values to obtain
I'll leave the response in kilojoules, rounded to two sig figs.
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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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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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