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Natali5045456 [20]
3 years ago
12

The specific heat of aluminum is 0.0215 cal/g°c. if a 4.55 g sample of aluminum absorbs 2.55 cal of energy, by how much will the

temperature of the sample change?
Physics
1 answer:
lakkis [162]3 years ago
8 0

The formula we will use on this one is:

E = m C ΔT

where,

E = energy

m = mass of sample

C = specific heat

ΔT = change in temperature

 

Calculating for ΔT:

ΔT = E / m C

ΔT = 2.55 cal / (4.55 g * 0.0215 cal/g°C)

ΔT = 0.098 °C

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A gas occupies a volume of 1.0 m3 in a cylinder at a pressure of 120kPa. A piston compresses the gas until the volume is 0.25m3,
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Answer:

Approximately 480\; \rm kPa, assuming that this gas is an ideal gas.

Explanation:

  • Let V(\text{Initial}) and P(\text{Initial}) denote the volume and pressure of this gas before the compression.
  • Let V(\text{Final}) and P(\text{Final}) denote the volume and pressure of this gas after the compression.

By Boyle's Law, the pressure of a sealed ideal gas at constant temperature will be inversely proportional to its volume. Assume that this gas is ideal. By this ideal gas law:

\displaystyle \frac{P(\text{Final})}{P(\text{Initial})} = \frac{V(\text{Initial})}{V(\text{Final})}.

Note that in Boyle's Law, P is inversely proportional to V. Therefore, on the two sides of this equation, "final" and "initial" are on different sides of the fraction bar.

For this particular question:

  • V(\text{initial}) = 1.0\; \rm m^3.
  • P(\text{Initial}) = 120\; \rm kPa.
  • V(\text{final}) = 0.25\; \rm m^3.
  • The pressure after compression, P(\text{Final}), needs to be found.

Rearrange the equation to obtain:

\displaystyle P(\text{Final}) = \frac{V(\text{Initial})}{V(\text{Final})} \cdot P(\text{Initial}).

Before doing any calculation, think whether the pressure of this gas will go up or down. Since the gas is compressed, collisions between its particles and the container will become more frequent. Hence, the pressure of this gas should increase.

\begin{aligned}P(\text{Final}) &= \frac{V(\text{Initial})}{V(\text{Final})} \cdot P(\text{Initial})\\ &= \frac{1.0\; \rm m^{3}}{0.25\; \rm m^{3}} \times 120\; \rm kPa = 480\; \rm kPa\end{aligned}.

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The tires of a car support the weight of a stationary car. If one tire has a slow leak, the air pressure within the tire will decrease with time, the surface area between the tire and the road will increase with time, and the net force the tire exerts on the road will be constant with time.

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