Ask question

Ask question

All categories

3 years ago

8

a wooden block has a mass of 1.2 kg, a specific heat of 710, and is at a temperature of 25*C. what is the block's final temperat

ure if it's thermal energy increases by 2,130J?

1 answer:

mash [69]3 years ago

8 0

The final temperature of the block is $27.5^{\circ} C$

Explanation:

The amount of thermal energy Q supplied to a substance is related to the increase in temperature of the substance, $\Delta T$ , according to the equation

$Q=mC_s \Delta T$

where:

m is the mass of the substance

$C_s$ is the specific heat capacity of the substance

In this problem, we have:

m = 1.2 kg is the mass of the block

$Q = 2,130 J$ is the amount of energy supplied to the block

$C_s = 710 J/kg^{\circ}C$ is the specific heat capacity of the block

Solving for $\Delta T$ , we find the increase in temperature:

$\Delta T = \frac{Q}{m C_s}=\frac{2130}{(1.2)(710)}=2.5^{\circ}C$

And since the initial temperature was

$T_i = 25^{\circ}C$

The final temperature will be

$T_f = T_i + \Delta T = 25+2.5=27.5^{\circ} C$

Learn more about specific heat capacity:

brainly.com/question/3032746

brainly.com/question/4759369

#LearnwithBrainly

You might be interested in

Sally travels by car from one city to another. She drives for 26.0 min at 83.0 km/h, 52.0 min at 41.0 km/h, and 45.0 min at 60.0

Anna007 [38]

The average speed is determined by the following formula:

average speed = [sum of (speed * time for which that speed was traveled)] / total time

average speed = [(83 * 26 + 41 * 52 + 60 * 45 + 0 * 15) / 60] / [(26 + 52 + 45 + 15) / 60]
*note: The division by 60 is to convert minutes to hours. We see that the 60 cancels from the top and bottom of the division

average speed = 50.65 km/hr

The total distance traveled is equivalent to the numerator of the fraction we used in the first part. This is:
Distance = (83 * 26 + 41 * 52 + 60 * 45 + 0 * 15) / 60

Distance = 116.5 kilometers

6 0

4 years ago

1. Set frequency of wave generator to 5Hz.

ivann1987 [24]

Separate the barriers so they have a 2cm gap between them.

3 0

3 years ago

If only 10 pounds is required to lift a 500-lb block, how much chain must be played out to lift the engine 3.0 inches?

adelina 88 [10]

Answer:

150 inches (12.5 ft)

Explanation:

The work done to lift the 500 pound block 3 inches should be the same as that to lift the 10 pond object a given distance.

The following is the equation one needs to solve:

$10 \,lb\,* \,x\,=\,500\,lb\,*\,3\,in\\10 \,lb\,* \,x\,=\,1500\,lb\,in\\$

therefore solving for the distance "x" gives as the answer (in inches):

$10 \,lb\,* \,x\,=\,1500\,lb\,in\\x\,=\,\frac{1500\,lb\,in}{10\,lb} \\\\x\,=150\,in$

which can also be given in feet as: 12.5 ft

3 0

3 years ago

The sound level at a distance of 2.30 m from a source is 115 dB. At what distance will the sound level have the following values

Aleksandr [31]

Answer:

distance is 13 m for 100 dB

distance is 409 km for 10 dB

Explanation:

Given data

distance r = 2.30 m

source β = 115 dB

to find out

distance at sound level 100 dB and 10 dB

solution

first we calculate here power and intensity and with this power and intensity we will find distance

we know sound level β = 10 log(I/ $I_{0}$ ) ......................a

put here value (I/ $I_{0}$ ) = 10^−12 W/m² and β = 115

115 = 10 log(I/10^−12)

so

I = 0.316228 W/m²

and we know power = intensity × 4π r² ...............b

power = 0.316228 × 4π (2.30)²

power = 21.021604 W

we know at 100 dB intensity is 0.01 W/m²

so by equation b

power = intensity × 4π r²

21.021604 = 0.01 × 4π r²

so by solving r

r = 12.933855 m = 13 m

distance is 13 m

and

at 10 dB intensity is 1 × 10^–11 W/m²

so by equation b

power = intensity × 4π r²

21.021604 = 1 × 10^–11 × 4π r²

by solving r we get

r = 409004.412465 m = 409 km

5 0

4 years ago

A mass of 100 g stretches a spring 5 cm. If the mass is set in motion from its equilibrium position with a downward velocity of

Rudiy27

Answer:

Explanation:

Given

mass of spring $m=100\ gm$

extension in spring $x=5\ cm$

downward velocity $v=70\ cm/s$

Position in undamped free vibration is given by

$u(t)=A\cos \omega _0t+B\sin \omega _0t$

where $\omega _0^2=\frac{k}{m}$

also $\frac{k}{m}=\frac{g}{L}$

$\omega _0^2=\frac{k}{m}=\frac{9.8}{0.05}$

$\omega _0=14$

$u(t)=A\cos(14t)+B\sin(14t)$

it is given

$u(0)=0$

$u'(0)=70\ cm/s$

substituting values we get

$A=0$

$u(t)=B\sin (14t)$

$u'(t)=14B\cos (14t)$

$70=14B$

$B=\frac{10}{2}$

$B=5$

$u(t)=5\sin (14t)$

3 0

4 years ago