A5.00-g sample of liquid water at 25.0 °c is heated by the addition of 84.0 j of energy. the final temperature of the water is °c. the specific heat capacity of liquid water is 4.18 j/g-k.

Final Temperature = 29.2 °C

Explanation:

Before moving any further, let's bring out the information given.

Mass of Liquid water (M) = 5.00g

Initial temperature (T1) = 25 C + 273 = 298 K (Converting to kelvin temperature)

Heat = 84.0 J

FInal Temperature (T2) = ?

Specific Heat Capacity of water (C) =  4.18 J/gK

The formular relating all these parameters is given as;

H = M * C * (T2 -T1)

Making T2 subject of formular, we have;

T2 = (H/MC) + T1

T2 = [84/(5*4.18)] + 298

T2 = 4.02 + 298 = 302 K

Converting to °C, T2 = 302 - 273 = 29.2 °C

a. they must be certified by the avma.

explanation:

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