Question

A chunk of brass with mass 0.6 kg, with specific heat 380 J/(kg· K), and with temperature 400 K is placed in 0.6 kg of water, with specific heat 4186 J/(kg· K), and with temperature 280 K. The water and the brass are in an insulated container and come to thermal equilibrium. Find the final temperature in kelvins (closest answer).

AND

The asteroid Oumuamua observed this year in November traveled around the sun at a speed of 87,500 m/s. Its proper length is 180 m. How much does a stationary observer near the sun measure its length to be changed at this speed due to relativity? A positive answer means its length is measured to be increased and a negative answer means that its length is measured to be decreased. (In micrometers, closest answer.)

AND

A radioactive sample is measured to be emitting 200 alpha particles per second. 20 minutes later it is emitting only 126 alpha particles per second. The half life of the unstable nucleus that is emitting these particles is - (in minutes, closest answer)

Answer 1

a) mass of brass m₁ = 0.6 kg

Specific heat capacity of brass S₁ = 380 J/(kg•K)

Initial temperature of brass T₁ = 400 K

mass of water m₂ = 0.6 kg

Specific heat capacity of water S₂ = 4186 J/(kg•K)

Initial temperature of water T₂ = 280 K

Let the final temperature of the system be T, equating heat lost and heat gained,we have

m₁S₁(T₁ - T) = m₂S₂(T - T₂)

i.e. 0.6×380×(400-T) = 0.6×4186×(T-280)

i.e. T = 290 K

Final temperature T = 290 K

b) Proper length l₀ = 180 m

Speed v = 87500 m/s

c = 3×10⁸ m/s

Now relativistic length l = y×l₀

Where y = 1/√(1-(v²/c²))

i.e. y = 1/√(1-(87500²/(3×10⁸)²)) = 1.000000042534

So l = y×l₀ = 1.000000042534×180 = 180.000007656 m

Change in length ∆l = l - l₀ = 180.000007656-180 = 0.000007656 m

So, ∆l = +(7.656 ×10^-6) m

c) From exponential decay law, we have I = I₀×e^-wt

Where I₀ = initial activity of sample at t = 0

I = activity of sample at time t

w = decay constant

Given I₀ = 200 emission s^-1

I = 126 emission s^-1

t = 20 min = 20×60=1200 s

Putting these values in exponential decay equation,

i.e. 126 = 200×e^-1200w

i.e. 1200w = -ln(126/200)

i.e. w = 0.000385 s^-1

Half life of sample = t_½ = (ln2)/w = (ln2)/(0.000385)

i.e. t_½ = 1800 s = 30 min