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irinina [24]
2 years ago
6

A hot metal bolt of mass 0.06 kg and specific heat capacity 899 J/kg°C is dropped into a calorimeter containing 0.16 kg water at

20°C. The bolt and water reach a final temperature of 25°C.
What is the initial temperature of the bolt?
Physics
1 answer:
skad [1K]2 years ago
7 0

Answer:

Explanation:

Given

mass of hot metal m_1=0.06\ kg

specific heat c=889\ J/kg-^{\circ}C

mass of water m_2=0.16\ kg

Temperature pf water T_2=20^{\circ}C

Final Temperature T=25^{\circ}C

let T_1 be the temperature of hot metal ball

Heat lost by heat metal bolt is gained by water in calorimeter

Heat lost by hot metal bolt Q_1=m\times c\times \Delta T

Q_1=0.06\times 889\times (T-25)

Heat gained by water Q_2=m_2\times c_w\times \Delta T

Q_2=0.16\times 4184\times (25-20)

Q_1=Q_2

0.06\times 889\times (T-25)=0.16\times 4184\times (25-20)

T_1=25+62.75

T_1=87.75^{\circ}C

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A positive kaon (K+) has a rest mass of 494 MeV/c² , whereas a proton has a rest mass of 938 MeV/c². If a kaon has a total energ
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Answer:

<em>0.85c </em>

Explanation:

Rest mass of Kaon M_{0K} = 494 MeV/c²

Rest mass of proton M_{0P}  = 938 MeV/c²

The rest energy is gotten by multiplying the rest mass by the square of the speed of light c²

for the kaon, rest energy E_{0K} = 494c² MeV

for the proton, rest energy E_{0P} = 938c² MeV

Recall that the rest energy, and the total energy are related by..

E = γE_{0}

which can be written in this case as

E_{K} = γE_{0K} ...... equ 1

where E = total energy of the kaon, and

E_{0} = rest energy of the kaon

γ = relativistic factor = \frac{1}{\sqrt{1 - \beta ^{2} } }

where \beta = \frac{v}{c}

But, it is stated that the total energy of the kaon is equal to the rest mass of the proton or its equivalent rest energy, therefore...

E_{K} = E_{0P} ......equ 2

where E_{K} is the total energy of the kaon, and

E_{0P} is the rest energy of the proton.

From E_{K} = E_{0P} = 938c²    

equ 1 becomes

938c² = γ494c²

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1.89\sqrt{1 - \beta ^{2} } = 1

squaring both sides, we get

3.57( 1 - \beta^{2}) = 1

3.57 - 3.57\beta^{2} = 1

2.57 = 3.57\beta^{2}

\beta^{2} = 2.57/3.57 = 0.72

\beta = \sqrt{0.72} = 0.85

but, \beta = \frac{v}{c}

v/c = 0.85

v = <em>0.85c </em>

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