0.90J/g°C
We are given:
Mass of Aluminium = 10 g
Quantity of heat = 677 Joules
Change in temperature = 125°C - 50°C
= 75°C
We are required to calculate the specific heat capacity of Aluminium
But, Quantity of heat = Mass × specific heat × Change in temperature
Q = mcΔt
Rearranging the formula;
c = Q ÷ mΔt
= 677 J ÷ (10 g × 75°C)
= 677 J ÷ 750g°C
= 0.903 J/g°C
= 0.90J/g°C
Thus, the specific heat capacity of Aluminium is 0.90J/g°C
Answer:
We are asked to describe the half-life of the radioactive materials.
63 mL
Explanation:
To find the amount of mililiters in an amount of liters, we must multiply by the amount of liters by 1000.
0.063 L × 1000 = 63 mL
Solid
C,B,(D or A)
