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butalik [34]
4 years ago
13

A calorimeter is a container that is insulated from the outside, so a negligible amount of energy enters or leaves the container

when it is closed.
Consider a copper calorimeter with mass 0.100 kg that contains 0.160 kg of water and 0.018 kg of ice, all in thermal equilibrium at atmospheric pressure.If 0.750 kg of lead at a temperature of 255 ∘C is dropped into the calorimeter, what is the final temperature?
Assume that no heat is lost to the surroundings.
Physics
1 answer:
Len [333]4 years ago
3 0

Answer:

This is a heat balance question with negligible heat losses  to the surroundings.

Sum of heat losses = Sum of the heat gains

We are required to find the final temperature of the calorimeter system T1?

we were given

C = specific heat capacities of Copper c, Water w, Ice i and Lead l

m = masses as mc = 0.1kg, mw = 0.16kg, mi=0.018kg, ml=0.75kg,

T = Temperature of the components, Lead Tl= 225C and at ambient pressure, Temp. To of the water, ice and copper = 25C

Since, at thermal equilibrium after adding the Lead, the lead temperature Tl decreases while copper, water and ice temp decrease, so we have respective heat losses 'q' as,

qc + qw + qi + ql =0

which means that

(mc x Cc x ΔTC) + (mw x Cw x ΔTC) + (mi x Ci x ΔTC) + (ml x Cl x ΔTC) = 0

(0.1x390x (T1 - 25)) + (0.16x4190x (T1 - 25)) + (0.018x4190x (T1 - 0)) + (0.750x130x (T1 - 225)) = 0

882.32T1 - 41558 = 0

T1 = 47.1K

Hence, the final temperature of the calorimeter system is 47.1K

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OverLord2011 [107]
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3 0
3 years ago
The temperature of water in a beaker is 45°C. What does this measurement represent?
julia-pushkina [17]

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8 0
3 years ago
Read 2 more answers
a cement block accidentally falls from rest from the ledge of a 80.6-m-high building. When the block is 10.8 m above the ground,
fomenos

Answer:

0.229 seconds

Explanation:

Given:

y₀ = 80.6 m

v₀ = 0 m/s

a = -9.8 m/s²

We need to find the difference in times when y = 10.8 m and y = 2.10 m.

When y = 10.8 m:

y = y₀ + v₀ t + ½ at²

10.8 = 80.6 + (0) t + ½ (-9.8) t²

10.8 = 80.6 − 4.9 t²

4.9 t² = 69.8

t = 3.774

When y = 2.10 m:

y = y₀ + v₀ t + ½ at²

2.10 = 80.6 + (0) t + ½ (-9.8) t²

2.10 = 80.6 − 4.9 t²

4.9 t² = 78.5

t = 4.003

The difference is:

4.003 − 3.774 = 0.229

The man has 0.229 seconds to get out of the way.

5 0
4 years ago
On a spending spree in Malaysia, you buy an ox with a weight of 28.9 piculs in the local unit of weights: 1 picul = 100 gins, 1
Talja [164]

Answer:

1747.41 kg

Explanation:

If          1 piculs = 100 gins,

Then     28.9 piculs = (28.9 × 100) gins = 2890 gins.

Also,

If         1 gin = 16 tahils

then   2890 gins =( 16 × 2890 ) = 46240 tahils

Also,

If         1 tahil = 10 chees,

Then    46240 tahils = 46240 × 10 chees = 462400 chees.

Also,

If           1 chee = 10 hoons

Then      462400 chees =( 10 × 462400) hoons = 4624000 hoons.

Also,

If            1 hoon = 0.3779 g,

Then     4624000 hoons = 0.3779 × 4624000 = 1747409.6 g.

The mass in kg that i will declare in the shipping manifest = (1747409.6/100 )kg = 1747.41 kg

6 0
3 years ago
You're driving down the highway late one night at 20 m/s when a deer steps onto the road 39 m in front of you. Your reaction tim
miss Akunina [59]

Answer:

a) 10.8 m

b) 24.3 m/s

Explanation:

a)

  • In order to get the total distance traveled since you see the deer till the car comes to an stop, we need to take into account that this distance will be composed of two parts.
  • The first one, is the distance traveled at a constant speed, before stepping on the brakes, which lasted the same time that the reaction time, i.e., 0.5 sec.
  • We can find this distance simply applying the definition of average velocity, as follows:

       \Delta x_{1} = v_{1o} * t_{react} = 20 m/s * 0.5 s = 10 m (1)

  • The second part, is the distance traveled while decelerating at -11 m/s2, from 20 m/s to 0.
  • We can find this part using the following kinematic equation (assuming that the deceleration keeps the same all time):

       v_{1f} ^{2}  - v_{1o} ^{2} = 2* a* \Delta x  (2)

  • where v₁f = 0, v₁₀ = 20 m/s, a = -11 m/s².
  • Solving for Δx, we get:

       \Delta x_{2} = \frac{-(20m/s)^{2}}{2*(-11m/s)} = 18.2 m (3)

  • So, the total distance traveled was the sum of (1) and (3):
  • Δx = Δx₁ + Δx₂ = 10 m + 18.2 m = 28.2 m (4)
  • Since the initial distance between the car and the deer was 39 m, after travelling 28.2 m, the car was at 10.8 m from the deer when it came to a complete stop.

b)

  • We need to find the maximum speed, taking into account, that in the same way that in a) we will have some distance traveled at a constant speed, and another distance traveled while decelerating.
  • The difference, in this case, is that the total distance must be the same initial distance between the car and the deer, 39 m.
  • ⇒Δx = Δx₁ + Δx₂ = 39 m. (5)
  • Δx₁, is the distance traveled at a constant speed during the reaction time, so we can express it as follows:

       \Delta x_{1} = v_{omax} * t_{react} = 0.5* v_{omax} (6)

  • Δx₂, is the distance traveled while decelerating, and can be obtained  using (2):

        v_{omax} ^{2} = 2* a* \Delta x_{2} (7)

  • Solving for Δx₂, we get:

       \Delta x_{2} = \frac{-v_{omax} ^{2} x}{2*a}  = \frac{-v_{omax} ^{2}}{(-22m/s2)} (8)

  • Replacing (6) and (8) in (5), we get a quadratic equation with v₀max as the unknown.
  • Taking the positive root in the quadratic formula, we get the following value for vomax:
  • v₀max = 24.3 m/s.
6 0
3 years ago
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