What 3 variables were changed in the pendulum experiment?

Chemistry

1 answer:

Paraphin [41]2 years ago

5 0

Answer:

In this experiment, the period of the pendulum is the dependent variable. There are three independent variables: the pendulum mass, the amplitude of the swing, and the length of the pendulum string.

Explanation:

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What happens to a water bottle in a freezer and why

Andrews [41]

The water well begin to freeze and turn into soild. why? because the temperature of the the freezer is getting the water bottle cold so when that cold temperature hits that that bottle which has liquid in it so you should know that when liquid is into cold air it well so be soild so when you put the water bottle in the freezer it will be soild.
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5 0

2 years ago

Which of the following substances (with specific heat capacity provided) would show the greatest temperature change upon absorbi

JulijaS [17]

Answer:

Pb is the substance that experiments the greatest temperature change.

Explanation:

The specific heat capacity refers to the amount of heat energy required to raise in 1 degree the temperature of 1 gram of substance. The highest the heat capacity, the more energy it would be required. These variables are related through the equation:

Q = c . m . ΔT

where,

Q is the amount of heat energy provided (J)

c is the specific heat capacity (J/g.°C)

m is the mass of the substance

ΔT is the change in temperature

Since the question is about the change in temperature, we can rearrange the equation like this:

$\Delta T = \frac{Q}{c.m}$

All the substances in the options have the same mass (m=10.0g) and absorb the same amount of heat (Q=100.0J), so the change in temperature depends only on the specific heat capacity. We can see in the last equation that they are inversely proportional; the lower c, the greater ΔT. Since we are looking for the greatest temperature change, It must be the one with the lowest c, namely, Pb with c = 0.128 J/g°C. This makes sense because Pb is a metal and therefore a good conductor of heat.

Its change in temperature is:

$\Delta T = \frac{q}{c.m} = \frac{100.0 J}{0.128 J/g.C . 10.0g } = 78.1$