Ask question

Ask question

All categories

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

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

You might be interested in

A kayaker moves 22 meters northward, then 18 meters southward , and finally 24 meters northward what is the distance and magnitu

Elina [12.6K]

Answer:

Distance = 64 metres

Displacement = 28 metres

Direction = northward

Explanation:

Given that a kayaker moves 22 meters northward, then 18 meters southward , and finally 24 meters northward

Distance covered is only about the magnitude of the length covered.

Distance = 22 + 18 + 24 = 64 metres

The direction will be considered when calculating the displacement

Let northward be positive and southward be negative

Displacement = 22 - 18 + 24 = 28 metres

Since the displacement is positive, the direction of the motion is northward.

8 0

3 years ago

Eats plants and meats

VMariaS [17]

Omnivores e<span>at both plants and meats. </span>

8 0

2 years ago

Read 2 more answers

Which two particles are present in the nucleus of an atom? Electrons and neutrons Electrons and molecules 1 TH O Protons and neu

aleksandrvk [35]

Answer:

Protons and neutrons

Explanation:

8 0

2 years ago

A cold beverage can be kept cold even a warm day if it is slipped into a porous ceramic container that has been soaked in water.

Arisa [49]

Answer:

The rate at which the container is losing water is 0.0006418 g/s.

Explanation:

Under the assumption that the can is a closed system, the conservation law applied to the system would be: $E_{in}-E_{out}=E_{change}$ , where $E_{in}$ is all energy entering the system, $E_{out}$ is the total energy leaving the system and, $E_{change}$ is the change of energy of the system.
As the purpose is to kept the beverage can at constant temperature, the change of energy ( $E_{change}$ ) would be 0.
The energy that goes into the system, is the heat transfer by radiation from the environment to the top and side surfaces of the can. This kind of transfer is described by: $Q=\varepsilon*\sigma*A_S*(T_{\infty}^4-T_S^4)$ where $\varepsilon$ is the emissivity of the surface, $\sigma=5.67*10^{-8}\frac{W}{m^2K}$ known as the Stefan–Boltzmann constant, $A_S$ is the total area of the exposed surface, $T_S$ is the temperature of the surface in Kelvin, $T_{\infty}$ is the environment temperature in Kelvin.
For the can the surface area would be ta sum of the top and the sides. The area of the top would be $A_{top}=\pi* r^2=\pi(0.0252m)^2=0.001995m^2$ , the area of the sides would be $A_{sides}=2*\pi*r*L=2*\pi*(0.0252m)*(0.09m)=0.01425m^2$ . Then the total area would be $A_{total}=A_{top}+A_{sides}=0.01624m^2$
Then the radiation heat transferred to the can would be $Q=\varepsilon*\sigma*A_S*(T_{\infty}^4-T_S^4)=1*5.67*10^{-8}\frac{W}{m^2K}*0.01624m^2*((32+273K)^4-(17+273K)^4)=1.456W$ .
The can would lost heat evaporating water, in this case would be $Q_{out}=\frac{dm}{dt}*h_{fg}$ , where $\frac{dm}{dt}$ is the rate of mass of water evaporated and, $h_{fg}$ is the heat of vaporization of the water ( $2257\frac{J}{g}$ ).
Then in the conservation balance: $Q_{in}-Q_{out}=Q_{change}$ , it would be $1.45W-\frac{dm}{dt}*2257\frac{j}{g}=0$ .
Recall that $1W=1\frac{J}{s}$ , then solving for $\frac{dm}{dt}$ : $\frac{dm}{dt}=\frac{1.45\frac{J}{s} }{2257\frac{J}{g} }=0.0006452\frac{g}{s}$

5 0

2 years ago

A particle having a speed of 0.87c has a momentum of 10-16 kg·m/s. what is its mass?

cupoosta [38]

The momentum p of a moving particle is the product between its mass, m, and tis velocity, v:
$p=mv$
In our problem, we know $p=10^{-16}~Kg \cdot m/s$ and $v=0.87c=0.87\cdot 3\cdot10^8~m/s=2.61\cdot 10^8~m/s$ , and using the relationship mentioned above, we can find the mass m of the particle:
$m= \frac{p}{v} =3.8\cdot10^{-25}~Kg$

8 0

3 years ago