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Vilka [71]
3 years ago
15

You are walking on a beach and find a mystery piece of metal. You take it back to your lab and measure its mass to be 0.5 kg. In

order to heat the metal 1 degree K, you must add 115 J of heat. Identify the metal. q =mCΔT
A) Brass
B) Copper
C) Platinum
D) Silver
Physics
1 answer:
murzikaleks [220]3 years ago
7 0

D) Silver

Explanation:

Given parameters:

Mass of metal = 0.5kg

temperature change = 1K

Quantity of heat = 115J

Unknown;

Identity of the metal = ?

Solution:

We first find the specific heat of the unknown metal and compare with the table of specific heat of metals.

   C = \frac{Q}{mΔt}

  C = \frac{115}{1 x 0.5} = 230JKg⁻¹K⁻¹

Comparing this value with the table shows that the metal is silver.

Learn more:

Specific heat brainly.com/question/7210400

#learnwithBrainly

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An unknown material, m1 = 0.45 kg, at a temperature of T1 = 91 degrees C is added to a Dewer (an insulated container) which cont
goldenfox [79]

Answer:

Explanation:

Let the specific heat of material be s

heat lost by material = m₁ s (T 1 - T ) ,  (T 1 - T ) is fall in temp , m₁ is mass of material

= .45 x s x (91 - 31.4 )

= 26.82 s

Heat gained by water

= m₂ cw (T2 - T )

1.3 x 4186 x ( 31.4 - 23 )

heat lost = heat gained

m₂ cw (T2 - T ) = m₁ s (T 1 - T )

1.3 x 4186 x ( 31.4 - 23 ) =  .45 x s x (91 - 31.4 )

45711.12 = 26.82 s

s = 1704.36

7 0
3 years ago
Can some help me, please?
cestrela7 [59]

Answer:

1. Transform Boundary

2.  Continental-continental convergent boundary

3. Oceanic-Oceanic Divergent boundary

Explanation:

8 0
3 years ago
Which statement describes a possible limitation in an experiment design?.... Apex
blondinia [14]
A ITS RIGHT I GOT 100 ON MY UNIT TEST
5 0
4 years ago
a banana boat accelerates from 4.167 m/s ar 2.00m/s^2.How far has it traveled when it reaches 8.333 m/s
Nataly_w [17]

Answer: The distance that we'll be travelled is d=16.35m

Explanation: The main idea here is to use the equation of motion. We are given the acceleration of 2m/s^2,the initial and final velocies. The formula to use is v^2=u^2+2as. Now we have to substitute the values.

2^2= 8.33^2+2(2)d

d= 16.35m

8 0
4 years ago
Suppose the collision between the packages is perfectly elastic. To what height does the package of mass m rebound?
Hitman42 [59]

The height, h to which the package of mass m bounces to depends on its initial velocity, v and the acceleration due to gravity, g and is given below:

h = \frac{v^{2}}{2g}

<h3>What are perfectly elastic collision?</h3>

Perfectly elastic collisions are collisions in which the momentum as well as the energy of the colliding bodies is conserved.

In perfectly elastic collisions, the sum of momentum before collision is equal to the momentum after collision.

Also, the sum of kinetic energy before collision is equal to the sum of kinetic energy after collision.

Since some of the Kinetic energy is converted to potential energy of the body;

\frac{mv^{2}}{2} = mgh

h = \frac{v^{2}}{2g}

Therefore, the height to which the package m bounces to depends on its initial velocity and the acceleration due to gravity.

Learn more about elastic collisions at: brainly.com/question/7694106

7 0
2 years ago
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