The energy change when the temperature of 0.750 kg of water decreased from 123 °C to 23.0 °C is -313.5 kJ.
In this problem, we are dealing with specific heat which is referred to as the sum of heat per unit mass required to raise the temperature by one degree Celsius.Since we know the boiling point of water is 100°C, at 123°C, the water is steam.
The formula we are referring to for calculating the heat energy Q
Q= m C ΔT
where m is the mass of water which is equal to 0.750 Kg, then
C is the specific heat of water equal to 4180 J/Kg °C and
ΔT is the change in temperature which is (23-123)= -100°C
So, assembling those desired values on the formula we get,
Q= 0.750 (4180) (-100)
Q=-313500
= -313.5 kJ
To know more about specific heat refer to the link brainly.com/question/13163208?referrer=searchResults.
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