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How many grams of water would require 2,153 joules of heat to raise its temperature from 44 °C to 81 °C? The specific heat of water is 4.184 J/g °C.

Answer 1

Isabella Alvarez

#"13.9 g"#

Use this equation

#"Q = mSΔT"#

Where

#"m" = "Q"/"SΔT"#

#color(white)("m") = "2153 J"/"4.184 J/g°C × (81 – 44)°C"#

#color(white)("m") = 13.9\ "g"#

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Answer 2

Adam Arnold

To calculate the amount of heat required to raise the temperature of water, you can use the formula:

Q = mcΔT

Where: Q = heat energy (in joules) m = mass of water (in grams) c = specific heat capacity of water (4.184 J/g°C) ΔT = change in temperature (in °C)

Given: ΔT = (81°C - 44°C) = 37°C c = 4.184 J/g°C Q = 2153 joules

Using the formula, we can rearrange it to solve for mass (m):

m = Q / (c * ΔT)

Plugging in the values:

m = 2153 J / (4.184 J/g°C * 37°C) m ≈ 14.66 grams

Therefore, approximately 14.66 grams of water would require 2153 joules of heat to raise its temperature from 44°C to 81°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.