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Par A

How much heat ) is absorbed by 9448 Dg of water in order for the temperature to increase from 25.00 C to 32.50 C7

asked
User Oshrat
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1 Answer

1 vote
To calculate the amount of heat absorbed by water to increase its temperature, we need to use the specific heat capacity of water and the formula:

Q = m * c * ΔT

where Q is the amount of heat absorbed, m is the mass of the water, c is the specific heat capacity of water, and ΔT is the change in temperature.

The specific heat capacity of water is 4.184 J/(g*C), which means that 1 gram of water requires 4.184 joules of energy to raise its temperature by 1 degree Celsius.

Substituting the given values, we have:

m = 9448 g (mass of water)
c = 4.184 J/(g*C) (specific heat capacity of water)
ΔT = (32.50 C - 25.00 C) = 7.50 C (change in temperature)

Q = m * c * ΔT
Q = 9448 g * 4.184 J/(g*C) * 7.50 C
Q = 283,464.72 J

Therefore, the amount of heat absorbed by 9448 g of water in order for the temperature to increase from 25.00 C to 32.50 C is approximately 283,465 J (joules).
answered
User Dja
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8.0k points
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