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iragen [17]
3 years ago
12

A 30.5 g sample of an alloy at 95.0°C is placed into 49.3 g water at 24.3°C in an insulated coffee cup. The heat capacity of the

coffee cup (without the water) is 9.2 J/K. If the final temperature of the system is 31.1°C, what is the specific heat capacity of the alloy? (c of water is 4.184 J/g×K)
Chemistry
1 answer:
user100 [1]3 years ago
4 0

Answer:

0.752 J/g*K

Explanation:

The heat lost by the alloy (which is negative) must be equal to the heat gained by the water and the coffee cup:

-Qa = Qw + Qc

-ma*ca*ΔTa = mw*cw*ΔTw + C*ΔTc

Where, m is the mass, c is the specific heat capacity, C is the heat capacity of the coffee cup, ΔT is the change in temperature, a represents the alloy, and w the water.

The coffee cup has initial temperature equal to the water, then:

-30.5*ca*(31.1 - 95.0) = 49.3*4.184*(31.1 - 24.3) + 9.2*(31.1 - 24.3)

1948.95ca = 1465.20

ca = 0.752 J/g*K

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two objects each have a mass of 5.0 g. one had a density of 2.7 g. and the other has a density of 8.4 g. which object has a larg
gulaghasi [49]

Answer:

The body with 2.7 \frac{g}{(cm^3)} has a larger volume

Explanation:

There is a missespelling in the question. The units of density cannot be g, they usually are \frac{g}{(cm^3)}

In order to answer this question, we need to use the formula of the density of a body:

Density = \frac{Mass}{Volume} = \frac{M}{V}

For body A:

Mass (M) = 5.0 g

Density (D) = 2.7 \frac{g}{(cm^3)}

2.7 \frac{g}{(cm^3)} = 5g * 1/V

V1 = 5g / (2.7 \frac{g}{(cm^3)})

V1 = 1.85 1.85 cm^3

For Body B:

Mass (M) = 5.0 g

Density (D) = 8.4 \frac{g}{(cm^3)}

8.4 \frac{g}{(cm^3)} = 5g * 1/V

V2 = 5g / (8.4 \frac{g}{(cm^3)})

V2 = 0.59 1.85 cm^3

So V1 is bigger than V2

5 0
3 years ago
Fission creates large amounts of: A.uranium B.air pollution C.carbon dioxide D.radioactive waste
Vaselesa [24]

Answer: -

Fission creates large amounts of radioactive waste.

Explanation: -

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Thus, Fission creates large amounts of radioactive waste.

5 0
4 years ago
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7. An 8.92 kilogram bar of steel at 25°C is heated in order to shape it into a beam. The amount of heat energy
kobusy [5.1K]

Answer:

1279 °C

Explanation:

Data Given:

Amount of Heat absorb = 5.82 x 10³ KJ

Convert KJ to J

1 KJ = 1000 J

5.82 x 10³ KJ = 5.82 x 10³ x 1000 = 5.82 x10⁶ J

mass of sample = 8.92 Kg

Convert Kg to g

1 kg = 1000 g

8.92 Kg = 8.92 x 1000 = 8920 g

Cs of steel = 0.51 J/g °C

change in temperature = ?

Solution:

Formula used

             Q = Cs.m.ΔT

rearrange the above equation to calculate the mass of steel sample

             ΔT =  Q / Cs.m  .... . . . . . (1)

Where:

Q = amount of heat

Cs = specific heat of steel = 0.51 J/g °C

m = mass

ΔT = Change in temperature

Put values in above equation 1

                ΔT = 5.82 x10⁶ J / 0.51  (J/g °C) x 8920 g

                ΔT = 5.82 x10⁶ J /4549.2 (J/°C)

                ΔT = 1279 °C            

So,

change in temperature = 1279 °C

4 0
3 years ago
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