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Delicious77 [7]
3 years ago
14

14. Heat energy equal to 25,000 J is applied to a 1200 g brick whose specific heat is 2.45 J/gºC.

Physics
1 answer:
choli [55]3 years ago
4 0

Answer:

∆T = 51 ºC

T2 = 76 ºC

Explanation:

(a)

∆Q = 25,000 J

m = 1200 g

c = 2.45 J/gºC

∆T = ?

∆Q = mc∆T

∆T = ∆Q/mc

∆T = 25000/(1200×2.45)

∆T = 51 ºC

(b) If T1 = 25 C

∆T = T2 - T1

T2 = ∆T + T1

T2 = 51+25

T2 = 76 ºC

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Dmitrij [34]

Explanation:

12) q = mCΔT

125,600 J = (500 g) (4.184 J/g/K) (T − 22°C)

T = 82.0°C

13) Solving for ΔT:

ΔT = q / (mC)

a) ΔT = 1 kJ / (0.4 kg × 0.45 kJ/kg/K) = 5.56°C

b) ΔT = 2 kJ / (0.4 kg × 0.45 kJ/kg/K) = 11.1°C

c) ΔT = 2 kJ / (0.8 kg × 0.45 kJ/kg/K) = 5.56°C

d) ΔT = 1 kJ / (0.4 kg × 0.90 kJ/kg/K) = 2.78°C

e) ΔT = 2 kJ / (0.4 kg × 0.90 kJ/kg/K) = 5.56°C

f) ΔT = 2 kJ / (0.8 kg × 0.90 kJ/kg/K) = 2.78°C

14) q = mCΔT

q = (2000 mL × 1 g/mL) (4.184 J/g/K) (80°C − 20°C)

q = 502,000 J

20) q = mCΔT

q = (2000 g) (4.184 J/g/K) (100°C − 15°C) + (400 g) (0.9 J/g/K) (100°C − 15°C)

q = 742,000 J

24) q = mCΔT

q = (0.10 g) (0.14 J/g/K) (8.5°C − 15°C)

q = -0.091 J

6 0
3 years ago
A merry go round (mass of 200kg and radius 5m) is rotating so that the outside edge is moving 10m/s. A giant bug of 6kd at a dis
romanna [79]

Answer:

2 rad/s

Explanation:

For a rotating object, the linear velocity is given by

v=\omega r

where \omega is the angular velocity and r is the radius.

\omega=\dfrac{v}{r}

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\omega=\dfrac{10 \text{ m/s}}{5 \text{ m}} = 2 \text{ rad/s}

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ad-work [718]

Answer:

B

Explanation:

using f=ma

a=f/m

=4000/2000

=2 m/s2

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2 years ago
Which of these is NOT an example of work?
amid [387]
The correct answer is A. Holding a heavy book.

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