Question

A 5.00-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.

Answer 1

q = m c (T2-T1)

84.0 J = 5.00g (4.18 J/gC) (T2 - 25.0 C)

T2 = 29.0 C

hope this helps!

(:

Answer 2

Final Temperature = 29.2 °C

Step-by-step 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