Question

how many grams of water will absorb a total of 2510 joules of energy during the temperature of the water changes from 10.0 degrees celsius to 35.0 degrees celsius?

Answer 1

ΔT = 30°C - 10 = 20°C.

1.0g of water will absorb 4.184Joules of heat energy per 1.0°C rise in temp.

2520J / (g x 4.184J/g/°C x 20°C ΔT).

Mass (g) = 2520 J / (4.184 J/g/°C x 20°C) = 2520 / 83.7 = c): 30 grams of water.

Step-by-step explanation:

Answer 2

24.02 grams of water will absorb a total of 2510 joules of energy during the temperature of the water changes from 10.0 degrees celsius to 35.0 degrees celsius.

Step-by-step explanation:

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

Between heat and temperature there is a direct proportionality relationship, where the constant of proportionality depends on the substance that constitutes the body and its mass. So, the equation that allows you to calculate heat exchanges is:

Q = c * m * ΔT

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

In this case,

• Q=2510 J
• water has a specific heat of about 4.18
  `(J)/(g*degrees celsius)`

• ΔT=35.0 degrees celsius- 10.0 degrees celsius= 25.0 degrees celsius

Replacing:

`2510 J= 4.18 (J)/(g*degrees celsius) *m*25.0 degrees celsius`

Solving you get:

`m=(2510 J)/(4.18 (J)/(g*degrees celsius) *25.0 degrees celsius)`

m≅ 24.02 g

24.02 grams of water will absorb a total of 2510 joules of energy during the temperature of the water changes from 10.0 degrees celsius to 35.0 degrees celsius.