Question

Calculate the amount of heat absorbed when 70 mL of water is heated from 30°C to 70°C. The specific heat of liquid water is 4.184 J/g °C. The density of water is about 1 g/mL. Report your answer in kJ to one decimal place.

Answer 1

Answer:

the heat absorbed by water is Q=11715.2 J

Explanation:

the amount of heat absorbed is

Q = m * c * (T - Ti)

where

Q= heat absorbed by water

m = mass f water heated

c = specific heat of water

T= final temperature , Ti= inicial temperature

also we know that the mass is related with the density through

density = mass/volume

d = m / V → m = d * V

replacing m in the heat equation

Q = m * c * (T - Ti) =  d * V * c * (T - Ti)

replacing values

Q = d * V * c * (T - Ti) = 1 g/ml * 70 ml * 4.184 J/g°C * (70°C - 30°C) = 11715.2 J

Note:

we assume

- constant density of the water between 30° and 70°

- constant specific heat of water between 30° and 70°

- the water has no impurities