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dem82 [27]
3 years ago
10

You pour 250 g of tea into a Styrofoam cup, initially at 80∘C and stir in a little sugar using a 100-g aluminum 20∘C spoon and l

eave the spoon in the cup. Assume the specific heat of tea is 4180 J/kg⋅∘C and the specific heat of aluminum is 900 J/kg⋅∘C. Part APart complete What is the highest possible temperature of the spoon when you finally take it out of the cup?
Physics
1 answer:
jok3333 [9.3K]3 years ago
3 0

To solve this problem it is necessary to apply the concepts related to the conservation of energy and heat transferred in a body.

By definition we know that the heat lost must be equal to the heat gained, ie

Q_g = Q_l

Where,

Q = Heat exchange

The heat exchange is defined as

Q = c_p m \Delta T

Where,

c_p = Specific heat

m = mass

\Delta T= Change in Temperature

Therefore replacing we have that

Q_g = Q_l

c_{p-tea} m \Delta T =  c_{p-al} m \Delta T

Replacing with our values we have that

0.25*4180*(80-T) = 0.1*900*(T-20)

11.61*(80-T) = T-20

T= \frac{948.8}{11.61}

T = 75.24\°C

Therefore the highest possible temperature of the spoon when you finally take it out of the cup is 75.24°C

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A ball with a mass of 2000 g is floating on the surface of a pool of water. What is the minimum volume that the ball could have
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2000\; {\rm cm^{3}}.

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The buoyancy force on this ball would be equal in magnitude to the weight of water that this ball has displaced.

Let m(\text{ball}) denote the mass of this ball. Let m(\text{water}) denote the mass of water that this ball has displaced.

Let g denote the gravitational field strength. The weight of this ball would be m(\text{ball}) \, g. Likewise, the weight of water displaced would be m(\text{water})\, g.

For this ball to stay afloat, the buoyancy force on this ball should be greater than or equal to the weight of this ball. In other words:

\text{buoyancy} \ge m(\text{ball})\, g.

At the same time, buoyancy is equal in magnitude the the weight of water displaced. Thus:

\text{buoyancy} = m(\text{water}) \, g.

Therefore:

m(\text{water})\, g = \text{buoyancy} \ge m(\text{ball})\, g.

m(\text{water}) \ge m(\text{ball}).

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\begin{aligned}V(\text{water}) &= \frac{m(\text{water})}{\rho(\text{water})} \\ &\ge \frac{m(\text{ball}))}{\rho(\text{water})}  \end{aligned}.

Given that m(\text{ball}) = 2000\; {\rm g} while \rho = 1.00\; {\rm g\cdot cm^{-3}}:

\begin{aligned}V(\text{water}) &\ge \frac{m(\text{ball}))}{\rho(\text{water})}  \\ &= \frac{2000\; {\rm g}}{1.00\; {\rm g\cdot cm^{-3}}} \\ &= 2000\; {\rm cm^{3}}\end{aligned}.

In other words, for this ball to stay afloat, at least 2000\; {\rm cm^{3}} of the volume of this ball should be under water. Therefore, the volume of this ball should be at least 2000\; {\rm cm^{3}}\!.

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