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stepan [7]
3 years ago
11

A 150 g ceramic serving bowl is warmed 52.0°C when it absorbs 2.2 kcal of heat from a serving of hot food. What is the specific

heat of the china dish?
Physics
1 answer:
allochka39001 [22]3 years ago
8 0

Answer:

1.18010256J/Kg°C

Explanation:

The quantity of heat energy is expressed as:

Q = mct

c is the specific heat capacity

t is change in temperature

Q = 4.184×2.2

Q = 9.2048Joules

m = 150g = 0.15kg

t = 52°C

Substitute and get c

9.2048 = 0.15c(52)

9.2048 = 7.8c

c = 9.2048/7.8

c = 1.18010256

Hence the specific heat capacity of the dish is 1.18010256J/Kg°C

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F_{mag}= BIL

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B = Magnetic Field

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L = Length

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F_{mag} = F_g

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Radius (r) = \frac{d}{2} = \frac{1*10^{-3}}{2} = 0.5*10^{-3}m

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From the relationship of density another way of expressing mass would be

\rho = \frac{m}{V} \rightarrow m = \rho V

At the same time the volume ratio for a cylinder (the shape of the wire) would be

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Re-arrange to find I

I = \frac{( \rho \pi r^2 L)g}{BL}

I = \frac{( \rho \pi r^2 )g}{B}

We have for definition that the Density of copper is 8.9*10^3 Kg/m^3, gravity acceleration is 9.8m/s^2 and the values of magnetic field (B) and the radius were previously given, then:

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I = 1370.05A

The current is too high to be transported which would make the case not feasible.

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