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Aleonysh [2.5K]

1 year ago

6

An object with a heat capacity of 345J∘C experiences a temperature change from 88.0∘C to 45.0∘C. How much heat is released in th

is process, in joules?

1 answer:

pogonyaev1 year ago

5 0

Answer:

There is 148.35 Joules of heat is released in the process.

Explanation:

Given that,

Heat capacity of the object, $c=345J/^oC$

Initial temperature, $T_i=88^{\circ}C$

Final temperature, $T_f=45^{\circ}C$

We need to find the amount of heat released in the process. It is a concept of heat capacity. The heat released in the process is given by :

$Q=mc\Delta T$

Let the mass of the object is 10 g or 0.01 kg

So,

$Q=0.01\times 345\times (88-45)$

Q = 148.35 Joules

So, there is 148.35 Joules of heat is released in the process. Hence, this is the required solution.

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When 30 V is applied across a resistor it generates 600 W of heat: what is the magnitude of its resistance?

grandymaker [24]

Answer:

<h2>1.5 ohms</h2>

Explanation:

Power is expressed as P = V²/R

R = resistance

V = supplied voltage

Given P = 600W and V = 30V

R = V²/P

R = 30²/600

R = 900/600

R = 1.5ohms

magnitude of its resistance is 1.5ohms

3 0

1 year ago

A 125-g metal block at a temperature of 93.2 °C was immersed in 100. g of water at 18.3 °C. Given the specific heat of the metal

Nataly_w [17]

Answer:

34.17°C

Explanation:

Given:

mass of metal block = 125 g

initial temperature $T_i$ = 93.2°C

We know

$Q = m c \Delta T$ ..................(1)

Q= Quantity of heat

m = mass of the substance

c = specific heat capacity

c = 4.19 for H₂O in $J/g^{\circ}C$

$\Delta T$ = change in temperature

Now

The heat lost by metal = The heat gained by the metal

Heat lost by metal = $125\times 0.9\times (93.2-T_f)$

Heat gained by the water = $100\times 4.184\times(T_f -18.3)$

thus, we have

$125\times 0.9\times (93.2-T_f)$ = $100\times 4.184\times(T_f -18.3)$

$10485-112.5T_f = 418.4T_f - 7656.72$

⇒ $T_f = 34.17^oC$

Therefore, the final temperature will be = 34.17°C

6 0

2 years ago

A child is sliding a toy block (with mass = m) down a ramp. The coefficient of static friction between the block and the ramp is

tiny-mole [99]

Answer:

$F=mg(sin(\theta )-0.25 cos(\theta ))$

Explanation:

The free body diagram of the block on the slide is shown in the below figure

Since the block is in equilibrium we apply equations of statics to compute the necessary unknown forces

N is the reaction force between the block and the slide

For equilibrium along x-axis we have

$\sum F_{x}=0\\\\mgsin(\theta )-\mu N-F=0\\\therefore F=mgsin(\theta)-\mu N......(\alpha )\\Similarly\\\sum F_{y}=0\\\\N-mgcos(\theta )=0\\\therefore N=mgcos(\theta ).......(\beta )\\\\$

Using value of N from equation β in α we get value of force as

$F=mg(sin(\theta )-\mu cos(\theta ))$

Applying values we get

$F=mg(sin(\theta )-0.25 cos(\theta ))$

8 0

1 year ago

Read 2 more answers

a. For a spring-mass oscillator, if you double the mass but keep the stiffness the same, by what numerical factor does the perio

Katena32 [7]

Answer:

a) factor $b=\sqrt{2}$

b) factor $b=\frac{1}{2}$

c) factor $b=1$

d) factor $b=1$

Explanation:

Time period of oscillating spring-mass system is given as:

$T=\frac{1}{f}$

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

where:

$f=$ frequency of oscillation

$m=$ mass of the object attached to the spring

$k=$ stiffness constant of the spring

a) <u>On doubling the mass:</u>

New mass, $m'=2m$

<u>Then the new time period:</u>

$T'=2\pi\sqrt{\frac{m'}{k} }$

$T'=2\pi\sqrt{\frac{2m}{k} }$

$T'=\sqrt{2}\times 2\pi\sqrt{\frac{m}{k} } }$

$T'=\sqrt{2} \times T$

where the factor $b=\sqrt{2}$ as asked in the question.

b) On quadrupling the stiffness constant while other factors are constant:

New stiffness constant, $k'=4k$

<u>Then the new time period:</u>

$T'=2\pi\sqrt{\frac{m}{k'} }\\\\T'=2\pi\sqrt{\frac{m}{4k} }\\\\T'=\frac{1}{2} \times 2\pi\sqrt{\frac{m}{k} } }\\\\T'=\frac{1}{2} \times T$

where the factor $b=\frac{1}{2}$ as asked in the question.

c) On quadrupling the stiffness constant as well as mass:

New stiffness constant, $k'=4k$

New mas, $m'=4m$

<u>Then the new time period:</u>

$T'=2\pi\sqrt{\frac{m'}{k'} }\\\\T'=2\pi\sqrt{\frac{4m}{4k} }\\\\T'=1 \times 2\pi\sqrt{\frac{m}{k} } }\\\\T'=1 \times T$

where factor $b=1$ as asked in the question.

d) On quadrupling the amplitude there will be no effect on the time period because T is independent of amplitude as we can observe in the equation.

so, factor $b=1$

7 0

1 year ago

What is the internal energy (to the nearest joule) of 10 moles of Oxygen at 100 K?

kkurt [141]

Answer:

U = 12,205.5 J

Explanation:

In order to calculate the internal energy of an ideal gas, you take into account the following formula:

$U=\frac{3}{2}nRT$ (1)

U: internal energy

R: ideal gas constant = 8.135 J(mol.K)

n: number of moles = 10 mol

T: temperature of the gas = 100K

You replace the values of the parameters in the equation (1):

$U=\frac{3}{2}(10mol)(8.135\frac{J}{mol.K})(100K)=12,205.5J$

The total internal energy of 10 mol of Oxygen at 100K is 12,205.5 J

6 0

2 years ago