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How much heat, in Joules, is required to raise the temperature of 28.41 grams of water from an initial temperature of 18.16°C to a final temperature of 44.28°C?

User Mhlandry
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Answer:

Step-by-step explanation:

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

Q = mcΔT

Where:

Q is the heat energy in Joules

m is the mass of the water in grams

c is the specific heat capacity of water, which is 4.18 J/g°C

ΔT is the change in temperature in °C

Given:

m = 28.41 grams

ΔT = (44.28°C - 18.16°C) = 26.12°C

Now, let's substitute these values into the formula:

Q = (28.41 g) x (4.18 J/g°C) x (26.12°C)

Calculating this gives us:

Q = 3039.4936 J

Therefore, the amount of heat required to raise the temperature of 28.41 grams of water from an initial temperature of 18.16°C to a final temperature of 44.28°C is approximately 3039.49 Joules.

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