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DaniilM [7]
3 years ago
14

A 0.500 kg block of lead is heated from 295 K to 350. K. How much heat was absorbed by the lead? (express your answer to the nea

rest joule
Physics
1 answer:
katen-ka-za [31]3 years ago
6 0

Answer:

The amount of energy needed to change the temperature of a certain supstance is given in the equation:

Q = c • m • ∆T

-Q is the amount of energy in Joules

-c is the specific heat capacity, a constant value for each supstance that shows amount of energy needed to raise the temperature of 1 g of a supstance by 1 K. Its unis is J/g•K (for lead, c=0.129J/g•K)

-m is the mass of the given supstance in grams

-∆T is the change in the temperature in Kelvins. In this case it's 350-295=55K

So, now, let's calculate:

Q = 0.129J/g•K • 500g • 55K

Q = 3,547.5 J

Rounded to the nearest whole number that is 3,548 Joules

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The work done by friction was -4.5\times10^{5}\ J

Explanation:

Given that,

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