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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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2 years ago
A hockey stick of mass ms and length L is at rest on the ice (which is assumed to be frictionless). A puck with mass mp hits the
krek1111 [17]

Answer:

L = mp*v₀*(ms*D) / (ms + mp)

Explanation:

Given info

ms = mass of the hockey stick

uis = 0 (initial speed of the hockey stick before the collision)

xis = D (initial position of center of mass of the hockey stick before the collision)

mp = mass of the puck

uip = v₀ (initial speed of the puck before the collision)

xip = 0 (initial position of center of mass of the puck before the collision)

If we apply

Ycm = (ms*xis + mp*xip) / (ms + mp)

⇒  Ycm = (ms*D + mp*0) / (ms + mp)

⇒  Ycm = (ms*D) / (ms + mp)

Now, we can apply the equation

L = m*v*R

where m = mp

v = v₀

R = Ycm

then we have

L = mp*v₀*(ms*D) / (ms + mp)

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