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lapo4ka [179]
3 years ago
8

Calculate the amount of heat transferred when 710 grams of water warms from an initial temperature of 4.0 ºC to a final temperat

ure of 25.0 ºC. The specific heat capacity of liquid water is 4.184 J/g ºC.
Physics
1 answer:
tigry1 [53]3 years ago
3 0

Answer:

Q = 62383.44 Joules

Explanation:

Given that,

Mass of water, m = 710 gm

Initial temperature of water, T_i=4^{\circ} C

Final temperature of water, T_f=25^{\circ} C

The specific heat capacity of liquid water is, c=4.184\ J/g\ ^oC

Heat transferred is given by :

Q=mc(T_f-T_i)

Q=710\times 4.184\times (25-4)

Q = 62383.44 Joules

So, the amount of heat transferred is 62383.44 Joules. Hence, this is the required solution.

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A system undergoes a two-step process. In the first step, the internal energy of the system increases by 222 J when 150 J of wor
joja [24]

Answer:0 J

Explanation:

Given

For first step

change in internal Energy of the system is \Delta U_1=222 J

Work done on the system W_1=-150 J

For second step

change in internal Energy of the system is \Delta U_2=123 J

Work done on the system W_2=-195 J

Work done on the system is considered as Positive and vice-versa.

and from first law of thermodynamics

Q=\Delta U+W

for first step

Q_1=222-150=72 J

Q_2=123-195=-72 J

overall heat added=Q_1+Q_2

Q_{net}=72-72 =0

For overall Process Heat added is 0 J

8 0
3 years ago
A fixed 11.2-cm-diameter wire coil is perpendicular to a magnetic field 0.53 T pointing up. In 0.10 s , the field is changed to
Karolina [17]

Answer:

The average induced emf in the coil is 0.0286 V

Explanation:

Given;

diameter of the wire, d = 11.2 cm = 0.112 m

initial magnetic field, B₁ = 0.53 T

final magnetic field, B₂ = 0.24 T

time of change in magnetic field, t = 0.1 s

The induced emf in the coil is calculated as;

E = A(dB)/dt

where;

A is area of the coil = πr²

r is the radius of the wire coil = 0.112m / 2 = 0.056 m

A = π(0.056)²

A = 0.00985 m²

E = -0.00985(B₂-B₁)/t

E = 0.00985(B₁-B₂)/t

E = 0.00985(0.53 - 0.24)/0.1

E = 0.00985 (0.29)/ 0.1

E = 0.0286 V

Therefore, the average induced emf in the coil is 0.0286 V

3 0
3 years ago
Which sentence states Newton’s third law?
hammer [34]
<span>If two objects collide, each object exerts a force equal to and in the opposite direction of the other.</span>
3 0
2 years ago
Read 2 more answers
A bodybuilder deadlifts a 215 kg weight to a height of 0.90 m above the ground. If he deadlifts this weight 10 times in a span o
wariber [46]

A bodybuilder deadlifts 215 kg to a height of 0.90 m. If he deadlifts this weight 10 times in 45 s, the power exerted is 421 W (b.)

<h3>What is power?</h3>

In physics, power (P) is the work (W) done over a period of time.

  • Step 1. Calculate the work done by the bodybuilder each time.

The bodybuilder lifts a 215 kg (m) weight to a height of 0.90 m (h). Being the gravity (g) of 9.81 m/s², we can calculate the work done in each lift using the following expression.

W = m × g × h = 215 kg × 9.81 m/s² × 0.90 m = 1.9 × 10³ N

  • Step 2. Calculate the work done by the bodybuilder over 10 times.

W = 10 × 1.9 × 10³ N = 1.9 × 10⁴ N

  • Step 3. Calculate the power exerted by the bodybuilder.

The bodybuilder does a work of 1.9 × 10⁴ N in a 45-s span.

P = 1.9 × 10⁴ N/45 s = 421 W

A bodybuilder deadlifts 215 kg to a height of 0.90 m. If he deadlifts this weight 10 times in 45 s, the power exerted is 421 W (b.)

Learn more about power here: brainly.com/question/911620

#SPJ1

4 0
1 year ago
A boat has a mass of 4040 kg. Its engines generate a drive force of 4660 N due west, while the wind exerts a force of 880 N due
maxonik [38]

Answer:

Explanation:

Given:

Mass of the boat, m = 4040 kg

The driving force of engine, FB = 4660 N in west = + 4660 N

The force of wind, Fwi = 880 N in east = -880 N

The force of water, Fwa = 1400 N in east = -1400N

Total three forces are acting on the boat

Fnet= Fb+fwi+Fwa

Fnet= 4660 - 880 - 1400

Fnet= +2380N

Acceleration (a) = Force/mass

= 2380/4040

= 0.59m/s2

6 0
3 years ago
Read 2 more answers
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