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andre [41]
3 years ago
6

A beaker of negligible heat capacity contains 456 g of ice at -25.0°C. A lab technician begins to supply heat to the container a

t the rate of 1000 J/min. How long after starting will it take before the temperature starts to rise above 0°C
Physics
1 answer:
Scorpion4ik [409]3 years ago
4 0

Answer:

176 min

Explanation:

456 g = .456 kg

Specific heat of ice s = 2093 J kg⁻¹

Heat required to raise the temperature by 25 degree

= mass x specific heat x rise in temperature.

= .456 x 2093 x 25

=23860 J

Heat required to melt the ice to make water at zero degree

= mass x latent heat

= .456 x 334 x 10³

=152304 J

Total heat required = 152304 + 23860 = 176164 J .

Time Required = Heat required / rate of supply of heat

= 176164 / 1000

176.16 min

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Suppose that a siren makes a sound reaching 120 decibels, but a wolf’s howl only reaches 90 decibels. 120 dB is how many times l
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First, we need to calculate the intensity of both the sounds. The formula for sound level is given as:

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<u>FOR SOUND OF SIREN:</u>

L = 120 dB

I = I₁ = ?

Therefore,

120 = 10 log[I₁/10⁻¹²]

log[I₁/10⁻¹²] = (120)/10

log[I₁/10⁻¹²] = 12

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I₁ = (10¹²)(10⁻¹²)

I₁ = 1

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L = 90 dB

I = I₂ = ?

Therefore,

90 = 10 log[I₂/10⁻¹²]

log[I₂/10⁻¹²] = (90)/10

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I₂ = (10⁹)(10⁻¹²)

I₂ = 10⁻³

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I₁/I₂ = 1/10⁻³

I₁/I₂ = 10³

<u>I₁/I₂ = 1000</u>

<u>Thus, the sound of siren is 1000 times louder than the sound of wolf's howl.</u>

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