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?

Answer 1

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

#q = mcΔt#, where
#m = "500 g"# #c = 1.00color(white)(l) "cal°C"^(-1)"g"^(-1)# #ΔT = T_2 –T_1 = "30 °C – 20°C" = "10 °C"#
∴ #q = mcΔT = 500 color(red)(cancel(color(black)("g"))) ×1.00color(white)(l) "cal"color(red)(cancel(color(black)("°C"^(-1)"g"^(-1)))) × 10 color(red)(cancel(color(black)("°C"))) = "5000 cal"#
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Answer 2

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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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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