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Taya2010 [7]
2 years ago
7

What is the change in velocity if the final velocity is 80 mph and the initial velocity is 20 mph? Is the object speeding up or

slowing down?
Physics
1 answer:
Dimas [21]2 years ago
7 0
Answer: Δv=vf-vi=80mph-20mph=60mph
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An insulated thermos contains 106.0 cm3 of hot coffee at a temperature of 80.0 °C. You put in 11.0 g of ice cube at its melting
Andrei [34K]

Answer:

the final temperature is T f = 64.977 ° C≈ 65°C

Explanation:

Since the thermus is insulated, the heat absorbed by the ice is the heat released by the coffee. Thus:

Q coffee + Q ice = Q surroundings =0 (insulated)

We also know that the ice at its melting point , that is 0 °C ( assuming that the thermus is at atmospheric pressure= 1 atm , and has an insignificant amount of impurities ).

The heat released by coffee is sensible heat : Q = m * c * (T final - T initial)

The heat absorbed by ice is latent heat and sensible heat : Q = m * L + m * c * (T final - T initial)

therefore

m co * c co * (T fco - T ico) + m ice * L + m ice * c wat  * (T fwa - T iwa) = 0

assuming specific heat capacity of coffee is approximately the one of water c co = c wa = 4.186 J/g°C and the density of coffee is the same as water

d co = dw = 1 gr/cm³

therefore m co = d co * V co = 1 gr / cm³ * 106 cm³ = 106 gr

m co * c wat * (T f  - T ico) + m ice * L + m ice * c wat  * (T f - T iwa) = 0

m co * c wat * T f+ m ice * c wat  * T f  = m ice * c wat  * T iwa  + m co * c wat * Tico -m ice * L

T f  = (m ice * c wat  * T iwa  + m co * c wat * Tico -m ice * L ) /( m co * c wat * + m ice * c wat )

replacing values

T f = (11 g * 4.186 J/g°C * 0°C +  106 g * 4.186 J/g°C*80°C - 11 g * 334 J/gr) / ( 11 g * 4.186 J/g°C +  106 g * 4.186 J/g°C* ) = 64,977 ° C

T f = 64.977 ° C

7 0
3 years ago
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