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igomit [66]
3 years ago
13

The specific heat capacity of copper is three times the specific heat capacity of lead. Equal masses of copper and lead are heat

ed from room temperature to the temperature of boiling water. To achieve this, the amount of heat added to the copper is ___________ the amount of heat added to the lead.
Physics
1 answer:
kondor19780726 [428]3 years ago
4 0

Answer:

three times

Explanation:

Heat added is calculated according to the following equation

Q = mcΔT

where Q is the heat added

           c is the specific heat capacity (of lead for the purposes of this question)

           ΔT is the temperature change in Kelvin or degree Celsius

Q1 for heating lead = mc(373K-298K) = 75mc

Q2 for heating copper = m(3c)(373-298) = 225mc = 3(75mc) = 3Q1

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The potential difference across a resistor increases by a factor of 2. How does the current change? (Ohm's law: V = IR) A. It de
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The answer is C.

The question says the potential difference is what is changing, which means we're solving for V.

It tells us that potential difference increases by a factor of two, which just means V doubles.

With this info, we can pick some numbers, plug it into Ohms law and see what happens.

Here's an example where I just picked random numbers that are easy to work with:

V=I*R
10=I*5
I=2
Lets increase the potential difference (V) by a factor of two and see what happens to current:
V=I*R
20=I*5 (all I've done is double the potential difference from 10 to 20)
I=4

When we increase V by a factor of 2, I increases by a factor of 2. We went from I=2 to I=4.
We can increase V by factor of 2 again and see:
V=I*R
40=I*5
I=8

Okay, current just increased by a factor of 2 again when we increased the potential difference by a factor of 2.

It's always good to check work with alternate numbers, so here's one more set:
V=I*R
16=I*4
(remember, we know we're solving for V, so I'm just plugging in random numbers for I and R)
I=4
Increase V by factor of 2:
32=I*4
I=8

So, when we increase V (the potential difference) by a factor of 2, I (current) always increases by a factor of 2 as well.

Hope this helps!





3 0
3 years ago
Read 2 more answers
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