Physics: Alarge punch bowl holds 3.95 kg of lemonade (which essentially is water) at 20.0 degrees celsius. a 45 gram ice cube at -10.2 degrees celsius is placed in the lemonade. what is the final temperature of the system? (ignore any heat exchange with the bowl)

Alarge punch bowl holds 3.95 kg of lemonade (which

Ответ:

tamyahamlin02p6b7yt

02.01.2022

Physics

$19 ^{\circ}$

Explanation:

Given

mass of water (m_w) 3.95 kg

Temperature of water is $20^{\circ} C$

mass of ice (m_i) =45 gm

Temperature of ice $=-10.2 ^{\circ} C$

specific heat of water 4.18 kJ/kg-K

specific heat of ice 2.22 kJ/kg-K

Latent heat of fusion(L) $3.4\times 10^5 J/kg$

Heat absorb by water when ice completely converts to water

$Q=mc_{ice}\delta T+mL$

$Q=0.045\times 2.22\times (0+10.2)+0.045\times 3.4\times 100$

Q=1.01898+15.3=16.3189 KJ

thus temperature of water will be

$3.95\times 4.18\times (20-T)=16.3189$

$T=19.011\approx 19^{\circ} C$

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Ответ:

smithmorgan773p35885

12.09.2022

Physics

p = 1.2 kg-m/s

Explanation:

The question is, "An object of mass 0.5kg is moving along a rectilinear path with constant acceleration of 0.3m / s2. If it started from rest and the magnitude of its momentum in kg * m / s after 8s is".

Mass of the object, m = 0.5 kg

Acceleration of the object, a = 0.3 m/s²

We need to find the momentum after 8 seconds.

We know that,

$p=F\times t$

i.e.

p = mat

So,

$p=0.5\times 0.3\times 8\\\\p=1.2\ kg-m/s$

So, the momentum of the object is 1.2 kg-m/s.

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Alarge punch bowl holds 3.95 kg of lemonade (which essentially is water) at 20.0 degrees celsius. a 45 gram ice cube at -10.2 degrees celsius is placed in the lemonade. what is the final temperature of the system? (ignore any heat exchange with the bowl)

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