Help me please:

How much he is required to raise the temperature of 464g of lead from 120°C to 219°C?

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.

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question: ray weighs 100 kilograms, his wrestling partner, bobby, weighs 50 kilograms. if ray moves at 4 m s , how fast will bobby have to run at him to apply an equal amount of kinetic energy?

answer: 8 m/s he needs to run twice as fast to keep up so he has some work to do

explanation:   if the mass has units of kilograms and the velocity of meters per second, the kinetic energy has units of kilograms-meters squared per second squared.

question answered by

(jacemorris04)