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
To raise the temperature of 70 mL of water from 30 °C to 70 °C, 11.7 kJ of heat must be absorbed.
First, we will convert 70 mL of water to grams using its density (1 g/mL).
[tex]70 mL \times 1 g/mL = 70 g[/tex]
We want to calculate the amount of heat (q) required to raise the temperature of 70 g of water from 30 °C to 70 °C.
We will use the following expression.
[tex]q = c \times m \times \Delta T = \frac{4.184 J}{g\° C } \times 70 g \times (70\° C-30\° C) \times \frac{1kJ}{1000J} = 11.7kJ[/tex]
where,
To raise the temperature of 70 mL of water from 30 °C to 70 °C, 11.7 kJ of heat must be absorbed.
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