If the temperature of 1 gram of water changes from 22°C to 27°C, how many calories of heat were involved? How many joules?

Answer 1

To calculate the heat energy, use the formula Q = mcΔT, where Q is the heat energy, m is the mass, c is the specific heat capacity of water, and ΔT is the temperature change. Specific heat capacity of water is approximately 4.18 J/g°C.

Q = (1 g) * (4.18 J/g°C) * (27°C - 22°C)

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Answer 2

Luna Chen

#"5 cal, 20.92 J"#

Use this equation

#"Q = mSΔT"#

Where

#"Q" = 1 cancel"g" × 1 "cal"/(cancel"g"cancel"°C") × (27 - 22) cancel"°C" = "5 cal"#

#"Q" = 1 cancel"g" × 4.184 "J"/(cancel"g"cancel"°C") × (27 - 22) cancel"°C" = "20.92 J"#

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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.