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Bingel [31]
3 years ago
6

A beaker contains 450g of water at a temperature of 24C, The therman capacity of the beaker is negligible and no heat is gained

by, or lost to, the atmosphere. Calculate the mass of ice, initially at 0C, that must be mixed with the water so that the final temp of the contents of the beakers is at 8C.
Physics
1 answer:
valentina_108 [34]3 years ago
8 0
Given:

450 grams of water (Tw = 24 C) 
Thermal capacity of the beaker = negligible
Q = 0 
Tice = 0 C
Tfinal = 8 C

450g * (Tf - Tw) + m_ice * (Tice - Tw) = 0

450g * (8 - 24) + m_ice (0 - 24) = 0 

solve for m_ice
 

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Tommy runs around a track whose circumference is 400 meters. He runs a single lap in a time of 62 seconds. What is Tommy’s displ
bija089 [108]

Answer:

<h2>Angular Displacement 6.28 radians</h2>

Explanation:

for circular motion we are expected to solve for Angular Displacement it is measured in radian

Measurement of Angular Displacement.

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∅= s/r

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given that

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we have to solve for the radius

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400= 2*3.142*r

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8 0
2 years ago
find the x-component of this vector 18.4,0.250. Remember, angles are measured from the +x axis. Find the x-component and y-compo
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Answer:

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7 0
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Read 2 more answers
Suppose you pour 0.250 kg of 20.0°C water into a 0.600 kg aluminum pan off the stove with a temperature of 173°C. Assume that th
lapo4ka [179]

Answer:

T_f=5.0116^{\circ}C

Explanation:

Given:

  • mass of water, m_w=0.25\ kg
  • initial temperature of water, T_i_w=20^{\circ}C
  • initial temperature of pan, T_i_p=173^{\circ}C
  • mass of pan, m_p=0.6\ kg
  • mass of water evapourated, m_v=0.03\ kg
  • specific heat of water, c_w=4186\ J.kg^{-1}.K^{-1}
  • specific heat of aluminium pan, c_a=900\ J.kg^{-1}.K^{-1}
  • latent heat of vapourization, L=2256000\ J.kg^{-1}

<u>Using the equation of heat:</u>

<em>Here, initially certain mass of water is vapourised first and then the remaining mass of water comes in thermal equilibrium with the pan.</em>

m_p.c_a.(T_{ip}-T_f)=m_v.L+(m_w-m_v).c_w.(T_f-T_{iw})

0.6\times 900\times (173-T_f)=0.03\times 2256000+(0.25-0.03)\times 4186\times (T_f-20)

T_f=5.0116^{\circ}C

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