Question

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

Answer 1

Answer:

4.704625 kJ (or 4704.625 J) is the amount of heat absorbed when the temperature of 75 grams of water increases from 20.°C to 35°C.

Explanation:

The measurement and calculation of the amounts of heat exchanged by a body or system is called calorimetry.

In this way, there is a direct proportionality relationship between heat and temperature. The constant of proportionality depends on the substance that constitutes the body as its mass, and the product of specific heat is produced by the body's mass. So, the equation that allows to calculate heat exchanges is:

Q = c * m * T

Where Q is the heat exchanged for a body of mass m, constituted by a specific heat substance c and where T is the temperature variation.

In this case:

• The specific heat capacity (c) of water is 4,181 J / g°C
• m=75 g
• T=35°C - 20°C ⇒ T= 15°C

Then:

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

Q≅4704.625 J

Being 1 kJ = 1000 J:

Q=4.704625 kJ

Finally, 4.704625 kJ (or 4704.625 J) is the amount of heat absorbed when the temperature of 75 grams of water increases from 20.°C to 35°C