What is the amount of heat absorbed when the temperature of 75 grams of water increases from 20.°C to 35°C

Chemistry

2 answers:

castortr0y [1.5K]23 days ago

5 0

Answer:

4702.5 J/g*k

Explanation:

This value varies according to the state of water, whether solid, liquid, or gas. For my calculations, I treated it as a liquid: J=(75g)(4.18 J/g*k)(15K)

alisha [1.6K]23 days ago

4 0

Answer:

4.704625 kJ (or 4704.625 J) is the heat obtained when 75 grams of water's temperature rises from 20°C to 35°C.

Explanation:

The process of quantifying heat transfer between systems or bodies is termed calorimetry.

In this context, there is a linear relationship between heat and temperature change. The proportionality constant relies on the substance's properties, derived from the mass times the specific heat. The specific equation utilized for calculating heat transfer is:

Q = c * m * T

Where Q denotes the heat exchange for a mass m, with a specific heat c, and T indicating the temperature difference.

In this scenario:

The specific heat of water is 4,181 J/g°C
m=75 g
T=35°C - 20°C = 15°C

Thus:

$Q= 4.181 \frac{J}{g C} * 75 g *15 C$

Q≅4704.625 J

Since 1 kJ = 1000 J:

Q=4.704625 kJ

Ultimately, 4.704625 kJ (or 4704.625 J) reflects the heat absorbed when 75 grams of water's temperature shifts from 20°C to 35°C

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