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
