Answer :
q = mCpΔT —> Solve for Cp —> Cp = q/mΔT
q = -6700 J (negative because it released heat
m = 70 g
ΔT = Final temp - initial temp = 25 - 90 = -65°C
Cp = -6700 J / (70 g)(-65°C) = 1.47 J/g°C
q = -6700 J (negative because it released heat
m = 70 g
ΔT = Final temp - initial temp = 25 - 90 = -65°C
Cp = -6700 J / (70 g)(-65°C) = 1.47 J/g°C
1.47 J/g°C is the specific heat capacity of a 70 g sample of an unknown metal that releases 6700J of heat when it cools from 90 to 25 degrees C.
What is specific heat capacity?
The specific heat capacity is defined as the quantity of heat (J) absorbed per unit mass (kg) of the material when its temperature increases by 1 K (or 1 °C), and its units are J/(kg K) or J/(kg °C).
q = mCpΔT
Solve for Cp —> Cp = q/mΔT
q = -6700 J (negative because it released heat)
m = 70 g
ΔT = Final temp - initial temp = 25 - 90 = -65°C
Cp = (-6700 J) ÷ (70 g)(-65°C)
= 1.47 J/g°C
Hence, 1.47 J/g°C is the specific heat capacity of a 70 g sample of an unknown metal that releases 6700J of heat when it cools from 90 to 25 degrees C.
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