The answer from this questions is the letter
B. He should not use his opinions as evidence.
I think it’s 44.6 J, but I’m not to sure so hoped this helped /:).
Answer:
The specific heat of aluminium is 0.8792 J/g °C or 0.21 Cal/g °C
Explanation:
Step 1 : Write formule of specific heat
Q=mcΔT
with Q = heat transfer (J)
with m = mass of the substance
with c = specific heat ⇒ depends on material and phase ( J/g °C)
with ΔT = Change in temperature
For this case :
Q = 1680 Calories = 7033.824 J ( 1 calorie = 4.1868 J)
m = 100.0g
c= has to be determined
ΔT = 100 - 20 = 80°C
<u>Step 2: Calculating specific heat</u>
⇒ via the formule Q=mcΔT
7033.824 J = 100g * c * 80
7033.824 = 8000 *c
c = 7033.824 /8000
c = 0,879228 J/g °C
or 0.21 Cal / g°C
The specific heat of aluminium is 0.8792 J/g °C or 0.21 Cal/g °C
Answer:
364 K or 91°C
Explanation:
Applying,
V₁/T₁ = V₂/T₂................ Equation 1
Where V₁ = Initial Volume, V₂ = Final volume, T₁ = initial Temperature, T₂ = final Temperature.
make T₂ the subject of the equation,
T₂ = V₂T₁/V₁................. Equation 2
From the question,
Given: V₁ = 375 mL, V₂ = 500 mL, T₁ = 0.0°C = (273+0) K = 273 K
Substitute these values into equation 2
T₂ = (500×273)/375
T₂ = 364 K
T₂ = (364-273) °C = 91 °C