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A sample of an unknown substance has a mass of 0.465 kg. if 3,000.0 j of heat is required to heat the substance from 50.0°c to 1

hammer [34]

The specific heat of the substance will be 0.129 J/g°C.

<h3>What is specific heat capacity?</h3>

The amount of heat required to increase a substance's temperature by one degree Celsius is known as specific heat capacity.

Similarly, heat capacity is the relationship between the amount of energy delivered to a substance and the increase in temperature that results.

The given data in the problem is;

Q is the amount of energy necessary to raise the temperature = 3,000.0 j

M is the mass= 0.465 kg.

Δt is the time it takes to raise the temperature.=50°c

s stands for specific heat capacity=?

Mathematically specific heat capacity is given by;

$\rm Q= MC \triangle t \\\\ C = \frac{Q}{M\triangle t} \\\\ C = \frac{3000}{0.465 \triangle 50} \\\\ C =129.0 J/Kg^0C \\\\ C= 0.129 J/g^0C$

Hence the specific heat of the substance will be 0.129 J/g°C.

To learn more about the specific heat capacity refer to the link brainly.com/question/2530523

5 0

2 years ago

Use the collision theory to explain how increasing the temperature of a reaction will affect the rate of the reaction.

kirill115 [55]

Increasing the temperature causes the particles in the reaction to become kinetically excited, hitting one another in increasing frequency. Increased collision among means faster rate or reaction.

7 0

3 years ago

As the temperature increases, materials with

Andrews [41]

Answer:large

Explanation:

As the temperature increases, materials with large coefficients of linear expansion increases a lot in size

7 0

3 years ago

A small ball of mass 2.00 kilograms is moving at a velocity 1.50 meters/second. It hits a larger, stationary ball of mass 5.00 k

rewona [7]

The kinetic energy of the small ball before the collision is

   KE = (1/2) (mass) (speed)²

   = (1/2) (2 kg) (1.5 m/s)

   =    (1 kg) (2.25 m²/s²)

   =    2.25 joules.

Now is a good time to review the Law of Conservation of Energy:

   Energy is never created or destroyed.
   If it seems that some energy disappeared,
   it actually had to go somewhere.
   And if it seems like some energy magically appeared,
   it actually had to come from somewhere.

The small ball has 2.25 joules of kinetic energy before the collision.
If the small ball doesn't have a jet engine on it or a hamster inside,
and does not stop briefly to eat spinach, then there won't be any
more kinetic energy than that after the collision. The large ball
and the small ball will just have to share the same 2.25 joules.

3 0

3 years ago

A horizontal spring-mass system has low friction, spring stiffness 160 N/m, and mass 0.3 kg. The system is released with an init

anygoal [31]

Answer:

(a) 0.38 m

(b) 2.78 m/s

(c) 0.11 watt

Explanation:

mass, m = 0.3 kg

spring constant, K = 160 N/m

initial compression, d = 12 cm = 01.2 m

initial speed, u = 3 m/s

(a) Let the maximum stretch is y.

Use conservation of energy

Initial potential energy + initial kinetic energy = final potential energy

0.5 x K x d² + 0.5 x m x u² = 0.5 x K x y²

160 x 0.12 x 0.12 + 0.3 x 0.12 x 0.12 = 160 x y²

2.304 + 0.00432 = 160 y²

y = 0.38 m

y = 38 cm

(b) Let v is the maximum speed.

The speed is maximum when the stretch in the spring is zero, so by use of conservation of energy

Initial potential energy + initial kinetic energy = final kinetic energy

0.5 x K x d² + 0.5 x m x u² = 0.5 x m x v²

160 x 0.12 x 0.12 + 0.3 x 0.12 x 0.12 = 0.3 x v²

2.304 + 0.00432 = 0.3 v²

v = 2.78 m/s

(c) The time period of the spring mass system is given by

$T=2\pi\sqrt{\frac{m}{K}}$

$T=2\pi\sqrt{\frac{0.3}{160}}$

T = 0.272 second

Energy dissipated per cycle = 0.03 J

Power, P = 0.03 / 0.272 = 0.11 Watt

5 0

3 years ago