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A 20 g piece of metal with a specific heat of .30 is at a temperature of 20C. A 30 g piece of metal with a specific heat of 1.5 is added and the final temperature for both is 80C. What was the initial temperature of the metal that was added?

User Breno Perucchi
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2.8k points

2 Answers

19 votes
19 votes

Answer:

2x – 3y = 11 …………….. (i)

x – 2y = – 12 …………….. (ii)

(ii) ⇒ x = 2y – 12 ……. (iii)

Substitute value of x from (iii) in (i), we get

2(2y – 12) + 3y = 11

⇒ 4y – 24 + 3y = 11

⇒ 7y = 35

⇒ y = 5

Substituting value of y = 5 in equation (iii), we get

4 = 2(5) – 12 = 10 – 12 = – 2

Hence, x = – 2, y = 5 is the required solution

Now, 5 = – 2m + 3

⇒ 2m = 3 – 5

⇒ 2m = – 2

⇒ m = – 1.

User Bernardo Siu
by
2.8k points
13 votes
13 votes

Answer:

60c

Step-by-step explanation:

80c-20c=60c i think I don't know

User Elric
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2.9k points