Respuesta :
d.
Q= m·ΔT·C(specific heat)= n·M·ΔT·C=10,5mol·27 g/mol· 194,5(225-30,5)·0,9=49600J=49,6kJ.
Q= m·ΔT·C(specific heat)= n·M·ΔT·C=10,5mol·27 g/mol· 194,5(225-30,5)·0,9=49600J=49,6kJ.
The amount of heat required will be "49.6 KJ".
Given:
Mole,
- m = 10.5 moles
Specific heat,
- C = 0.900 j/g•°C
Temperature,
- T = 30.5°C - 225°C
As we know,
→ [tex]Q = m \Delta T C[/tex]
By substituting the values, we get
→ [tex]= 10.5\times 27\times 0.9\times (225-30.5)[/tex]
→ [tex]=255.15\times (194.5)[/tex]
→ [tex]= 49.6 \ KJ[/tex]
Thus response above is correct.
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