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4 years ago

12

How much energy is required to raise the temperature of one kilogram (liter) of water 1°c?

1 answer:

Gnoma [55]4 years ago

4 0

Heat required in a system can be calculated by multiplying the given mass to the specific heat capacity of the substance and the temperature difference. It is expressed as follows:<span>

Heat = mC(T2-T1)
Heat = 1 kg (4.18 kJ / kg C)( 1 C)
<span>Heat = 4.18 kJ energy needed</span></span>

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What is the acceleration of a 0.30-kg volleyball when a player uses a force of 42 N to spike the ball?

quester [9]

Answer:

The acceleration will be 140 meter per second

Explanation:

Force F = mass m × acceleration a

If F = 42 N and m = 0.30 kg

Then acceleration a = F/m

a = 42/0.30

a = 140 m/s

5 0

4 years ago

S.I unit for moment of inertia of a fly wheel

qaws [65]

Answer:

m is expressed in kilograms and r in metres, with I (moment of inertia) having the dimension kilogram-metre square.

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2 years ago

A 1.50 cm high diamond ring is placed 20.0 cm from a concave mirror with radius of curvature 30.00 cm. The magnification is ____

Rama09 [41]

Answer:

Magnification, m = -0.42

Explanation:

It is given that,

Height of diamond ring, h = 1.5 cm

Object distance, u = -20 cm

Radius of curvature of concave mirror, R = 30 cm

Focal length of mirror, f = R/2 = -15 cm (focal length is negative for concave mirror)

Using mirror's formula :

$\dfrac{1}{f}=\dfrac{1}{u}+\dfrac{1}{v}$ , f = focal length of the mirror

$\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}$

$\dfrac{1}{v}=\dfrac{1}{-15}+\dfrac{1}{-20}$

v = -8.57 cm

The magnification of a mirror is given by,

$m=\dfrac{-v}{u}$

$m=\dfrac{-(-8.57)}{-20}$

m = -0.42

So, the magnification of the concave mirror is 0.42. Thew negative sign shows that the image is inverted.

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4 years ago

You have to choose one of these Kpop members for the purge, or you can group them into three. Who do you choose, what are their

Burka [1]

Answer:

Wonho, Seonghwa, and Bangchan

Explanation:

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3 years ago

Read 2 more answers

Luis and Aisha conducted an experiment. They exerted different forces on four objects. Their results are shown in the table.

lesya692 [45]

Answer:

Object 3 has greatest acceleration.

Explanation:

Objects Mass Force

1 10 kg 4 N

2 100 grams 20 N

3 10 grams 4 N

4 1 kg 20 N

Acceleration of object 1,

$a_1=\dfrac{F_1}{m_1}\\\\a_1=\dfrac{4}{10}\\\\a_1=0.4\ m/s^2$

Acceleration of object 2,

$a_2=\dfrac{F_2}{m_2}\\\\a_2=\dfrac{20}{0.1}\\\\a_2=200\ m/s^2$

Acceleration of object 3,

$a_3=\dfrac{F_3}{m_3}\\\\a_3=\dfrac{4}{0.01}\\\\a_3=400\ m/s^2$

Acceleration of object 4,

$a_4=\dfrac{F_4}{m_4}\\\\a_4=\dfrac{20}{1}\\\\a_3=20\ m/s^2$

It is clear that the acceleration of object 3 is $400\ m/s^2$ and it is greatest of all. So, the correct option is (3).

4 0

3 years ago

Read 2 more answers