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2 years ago

What is the power produced by a light bulb with a voltage of 120 volts and a current of 0.045 amps? PLEASE HELP I HAVE 5 MINUTES

Chemistry

2 answers:

Artyom0805 [142]2 years ago

8 0

The answer is 5.4! I found it on a an online calculator!

GuDViN [60]2 years ago

7 0

Answer : The power produced by light bulb is, 5.4 watts

Solution : Given,

Voltage = 120 V

Current = 0.045 A

Formula used :

$P=V\times I$

where,

P = power produced by light bulb

V = voltage

I = current

Now put all the given values in the above formula, we get the power produced by the light bulb.

$P=(120V)\times (0.045A)=5.4watts$

Therefore, the power produced by light bulb is, 5.4 watts

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Vlad [161]

Explanation:

a) The amount of heat released by coffee will be absorbed by aluminium spoon.

Thus, $heat_{absorbed}=heat_{released}$

To calculate the amount of heat released or absorbed, we use the equation:

$Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})$

Also,

$m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]$ ..........(1)

where,

q = heat absorbed or released

$m_1$ = mass of aluminium = 45 g

$m_2$ = mass of coffee = 180 g

$T_{final}$ = final temperature = ?

$T_1$ = temperature of aluminium = $24^oC$

$T_2$ = temperature of coffee = $85^oC$

$c_1$ = specific heat of aluminium = $0.80J/g^oC$

$c_2$ = specific heat of coffee= $4.186 J/g^oC$

Putting all the values in equation 1, we get:

$45 g\times 0.80J/g^oC\times (T_{final}-24^oC)=-[180 g\times 4.186J/g^oC\times (T_{final}-83^oC)]$

$T_{final}=80.30^oC$

80.30 °C is the final temperature.

b) Energy flows from higher temperature to lower temperature.Whenever two bodies with different energies and temperature come in contact. And the resulting temperature of both bodies will less then the body with high temperature and will be more then the body with lower temperature.

So, is our final temperature of both aluminium and coffee that is 80°C less than initial temperature of coffee and more than the initial temperature of the aluminum.

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