Respuesta :
Answer: 1045 J of energy was released on cooling the down the water from 20 °C to 10 °C.
Explanation:
Specific heat of water : 4.18J/g °C(standard value)
[tex]Q=mc\Delta T=[/tex]
Q= Energy released
m = mass of the substance
c = specific heat value of a substance
[tex]\Delta T[/tex] = Change in temperature
For 25 g of water the Q released when cooled down to 20 °C to 10 °C will be:
[tex]Q=mc\Delta T=25 g\times 4.18J/g ^oC\times (10^oC-20^oC)=-1045 J[/tex]
Q = -1045 J (negative sign indicates that energy is released)
1045 J of energy was released on cooling the down the water from 20 °C to 10 °C.
The quantity of heat energy released is 1046 J
From the question,
We are to determine the heat energy released,
From the specific heat formula,
Q = mcΔT
Where Q is heat energy
m is the mass
c is the specific heat
ΔT is the change in temperature
From the question,
m = 25.0 grams
ΔT = (20.0 - 10.0) °C = 10.0 °C
c = 4.184 J/g °C (Specific heat capacity of water)
Putting the parameters into the equation
Q = mcΔT
We get
Q = 25.0 × 4.184 × 10.0
Q = 1046 J
Hence, the quantity of heat energy released is 1046 J
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