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

level 4 to level 2 : ))

answer: semi-infinite solid initially at a uniform temperature of 400 ◦c is cooled at its surface by convection. the cooling fluid at a constant temperature of 200 ◦c has a convective coefficient (ha) of 250 w/m2.k. the physical properties of the solid: k = 20 w/m.k and α = 4 * 10-5  m2/s. using δx = 0.04 m and δt = 10 s. write the finite difference equations describing the transient heat transfer of the first five δx and the temperature profile after 30 s.