How much heat is absorbed when a 10.0-g sample of water is changed from 23.0⁰C to 32.0⁰C?

376.56J

Explanation:

The amount or quantity of heat absorbed can be calculated using the formula:

Q = m × c × ∆T

Where; Q = amount of heat absorbed (J)

m = mass of water (g)

c = specific heat capacity of water (J/g°C)

∆T = change in temperature (°C)

According to the provided information in this question; m = 10.0g, c = 4.184J/g°C, final temperature = 32°C, initial temperature = 23°C

Q = m × c × ∆T

Q = 10 × 4.184 × (32 - 23)

Q = 41.84 × 9

Q = 376.56J

the density(ratio between mass and volume as m/v) decreases from top to bottom in periodic table group as the size increases and nuclear charge decreases. but in periods density increases from left to right as atom size decreases and nuclear charge increases.

the product of the pressure and volume is also constant