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:
Ionic compounds don't conduct electricity very well because the charge carriers can't move through the crystal. They can conduct heat because the kinetic energy itself is the "heat carrier”.
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It is , how close a measurement is to an accepted value for the measurement !
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1. Atomic Mass
2. Highest melting point
3. Configuration
4. A
5. A
6. D
7. A
8. C
9. B
10. A
11. C
12. C
13. A
Answer:25grams
Explanation: the mass stays the same even when it changes form
