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Fofino [41]
3 years ago
15

Use the table below to answer the following questions. Substance Specific Heat (J/g•°C) water 4.179 aluminum 0.900 copper 0.385

iron 0.450 granite 0.790 1. When 3.0 kg of water is cooled from 80.0C to 10.0C, how much heat energy is lost? 2. How much heat is needed to raise a 0.30 kg piece of aluminum from 30.C to 150C? 3. Calculate the temperature change when: a) 10.0 kg of water loses 232 kJ of heat. b) 1.96 kJ of heat are added to 500. g of copper. 4. When heated, the temperature of a water sample increased from 15°C to 39°C. It absorbed 4300 joules of heat. What is the mass of the sample? 5. 5.0 g of copper was heated from 20°C to 80°C. How much energy was used to heat Cu? 6. The temperature of a sample of iron with a mass of 10.0 g changed from 50.4°C to 25.0°C with the release of 47 Joules of heat. What is the specific heat of iron? 7. The temperature of a sample of water increases
Physics
1 answer:
lbvjy [14]3 years ago
8 0

1. -8.78 \cdot 10^5 J

The energy lost by the water is given by:

Q=m C_s \Delta T

where

m = 3.0 kg = 3000 g is the mass of water

Cs = 4.179 J/g•°C is the specific heat

\Delta T=10.0C-80.0C=-70.0 C is the change in temperature

Substituting,

Q=(3000 g)(4.179 J/gC)(-70.0 C)=-8.78 \cdot 10^5 J

2. 3.24 \cdot 10^4 J

The energy added to the aluminium is given by:

Q=m C_s \Delta T

where

m = 0.30 kg = 300 g is the mass of aluminium

Cs = 0.900 J/g•°C is the specific heat

\Delta T=150.0 C-30.0C =120.0 C is the change in temperature

Substituting,

Q=(300 g)(0.900 J/gC)(120.0 C)=3.24 \cdot 10^4 J

3a. -5.6^{\circ}C

The temperature change of the water is given by

\Delta T=\frac{Q}{m C_s}

where

Q = -232 kJ=-2.32\cdot 10^5 J is the heat lost by the water

m=10.0 kg=10000 g is the mass of water

Cs = 4.179 J/g•°C is the specific heat

Substituting,

\Delta T=\frac{-2.32\cdot 10^5 J}{(10000g)(4.179 J/gC)}=5.6^{\circ}C

3b. +10.2^{\circ}C

The temperature change of the copper is given by

\Delta T=\frac{Q}{m C_s}

where

Q = 1.96 kJ=1960 is the heat added to the copper

m= 500 g is the mass of copper

Cs = 0.385 J/g•°C is the specific heat

Substituting,

\Delta T=\frac{1960 J}{(500g)(0.385 J/gC)}=10.2^{\circ}C

4. 42.9 g

The mass of the water sample is given by

m=\frac{Q}{C_S \Delta T}

where

Q=4300 J is the heat added

\Delta T=39 C-15 C=24C is the temperature change

Cs = 4.179 J/g•°C is the specific heat

Substituting,

m=\frac{4300 J}{(4.179 J/gC)(24 C)}=42.9 g

5. 115.5 J

The heat used to heat the copper is given by:

Q=m C_s \Delta T

where

m = 5.0 g is the mass of copper

Cs = 0.385 J/g•°C is the specific heat

\Delta T=80.0 C-20.0C =60.0 C is the change in temperature

Substituting,

Q=(5.0 g)(0.385 J/gC)(60.0 C)=115.5 J

6. 0.185 J/g•°C

The specific heat of iron is given by:

C_s = \frac{Q}{m \Delta T}

where

Q = -47 J is the heat released by the iron

m = 10.0 g is the mass of iron

\Delta T=25.0-50.4 C=-25.4 C is the change in temperature

Substituting,

C_s = \frac{-47 J}{(10.0 g)(-25.4 C)}=0.185 J/gC

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If the absolute temperature of a gas is 600 K, the temperature in degrees Celsius is
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T(K) = T(C) + 273
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<span>The answer is letter B.</span>
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Which of the following electromagnetic waves has the shortest wavelength?
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Ultraviolet

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When the distance between two objects doubles, what happens to the gravitational attraction between these two objects?
Feliz [49]

Answer:

As indicated by Newton's law of attraction each article or body in the universe draws in every single item towards one another and that power of fascination is straightforwardly relative to the result of their masses and contrarily corresponding to the square of the distance between them.  

The power of gravity between two articles will diminish as the distance between them increments. The two most significant elements influencing the gravitational power between two items are their mass and the distance between their focuses. As mass increments, so does the power of gravity, however an increment in distance mirrors a reverse proportionality, which makes that power decline dramatically.  

At that point by Newton's All inclusive Law of Attractive energy;  

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A helicopter lifts a 72 kg astronaut 15 m vertically from the ocean by means of a cable. The acceleration of the astronaut is g/
NemiM [27]

Answer:

a) F_H=776.952\ N

b) F_g=706.32\ N

c) v=5.4249\ m.s^{-1}

d) KE=1059.48\ J

Explanation:

Given:

  • mass of the astronaut, m=72\ kg
  • vertical displacement of the astronaut, h=15\ m
  • acceleration of the astronaut while the lift, a=\frac{g}{10} =0.981\ m.s^{-2}

a)

<u>Now the force of lift by the helicopter:</u>

Here the lift force is the resultant of the force of gravity being overcome by the force of helicopter.

F_H-F_g=m.a

where:

  • F_H= force by the helicopter
  • F_g= force of gravity

F_H=72\times 0.981+72\times9.81

F_H=776.952\ N

b)

The gravitational force on the astronaut:

F_g=m.g

F_g=72\times 9.81

F_g=706.32\ N

d)

Since the astronaut has been picked from an ocean we assume her initial velocity to be zero, u=0\ m.s^{-1}

using equation of motion:

v^2=u^2+2a.h

v^2=0^2+2\times 0.981\times 15

v=5.4249\ m.s^{-1}

c)

Hence the kinetic energy:

KE=\frac{1}{2} m.v^2

KE=0.5\times 72\times 5.4249^2

KE=1059.48\ J

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