How many calories are absorbed by a pot of water with a mass of 500 g in order to raise the temperature from 20° C to 30° C?
The water absorbs 5000 cal.
It is my assumption that the pot does not absorb heat.
The equation for the heat that the water absorbs is
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To calculate the calories absorbed by water, we use the formula:
Q = m * c * ΔT
where: Q is the heat absorbed (in calories), m is the mass of the water (in grams), c is the specific heat capacity of water (1 calorie/gram °C), and ΔT is the change in temperature (final temperature - initial temperature).
Substituting the values: m = 500 g, c = 1 cal/g°C, ΔT = 30°C - 20°C = 10°C
Q = 500 g * 1 cal/g°C * 10°C = 5000 calories
Therefore, the pot of water absorbs 5000 calories to raise its temperature from 20°C to 30°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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