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4709.25 joules of heat are absorbed when 25 grams of water increases temperature by 45 K.

Let suppose that temperature change experienced by water is entirely sensible. Then, we can estimated the heat absorbed by the water sample ([tex]Q[/tex]), in joules, through this expression:

[tex]Q = m\cdot c \cdot \Delta T[/tex] (1)

Where:

  • [tex]m[/tex] - Mass of the water sample, in grams.
  • [tex]c[/tex] - Specific heat of water, in joules per kilogram-Kelvin.
  • [tex]\Delta T[/tex] - Temperature change, in Kelvin.

If we know that [tex]m = 25\,g[/tex], [tex]c = 4.186\,\frac{J}{g\cdot K}[/tex] and [tex]\Delta T = 45\,K[/tex], then the heat absorbed by the water sample is:

[tex]Q = (25\,g)\cdot \left(4.186\,\frac{J}{g\cdot K} \right)\cdot (45\,K)[/tex]

[tex]Q = 4709.25\,J[/tex]

4709.25 joules of heat are absorbed when 25 grams of water increases temperature by 45 K.

We kindly invite to check this question on sensible heat: https://brainly.com/question/11325154