Answer:
Recorded
Explanation:
If you don't record your information you may need to prove your work and you might forget it!
Answer:
24805.44 J
Explanation:
Step 1:
Data obtained from the question.
Mass (M) = 464g
Initial temperature (T1) = 120°C
Final temperature (T2) = 219°C
Change in temperature (ΔT) = T2 - T1 = 219°C - 120°C = 99°C
Specific heat capacity of lead (C) = 0.129cal/g°C = 4.184 x 0.129 = 0.54J/g°C
Heat (Q) =?
Step 2:
Determination of the heat Q, required the temperature of lead. This is illustrated below
Q = MCΔT
Q = 464 x 0.54 x 99
Q = 24805.44 J
Therefore, 24805.44 J of heat is required to raise the temperature of lead.
Answer:
2.107×10²⁴
Explanation:
no of moles=no of particles/Avogadro's number
3.5=X/6.02×10²³
by cross multiplication
3.5×6.02×10²³
=2.107 ×10²⁴
Answer:
Q = -811440 J
Explanation:
Given data:
Mass of oil = 2.76 Kg (2.76× 1000 = 2760 g)
Initial temperature = 191 °C
Final temperature = 23°C
Specific heat capacity of oil = 1.75 J/g.°C
Solution:
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
ΔT = 23°C - 191 °C
ΔT = -168°C
Q = 2760 g ×1.75 J/g.°C ×-168°C
Q = -811440 J
Negative sign show heat is released.
The third one because
Most minerals can be characterized and classified by their unique physical properties: hardness, luster, color, streak, specific gravity, cleavage, fracture, and tenacity.