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Dovator [93]
3 years ago
8

A 40.2 g sample of a metal heated to 99.3°C is placed into a calorimeter containing 120 g of water at 21.8°C. The final temper

ature of the water is 24.5°C. Which of the following might be the metal that was used?
A) Aluminum (c=0.89J/g°C)
B) Iron (c=0.45J/g°C)
C) Copper (0.20J/g°C)
D) Lead (c=0.14J/g°C)
Chemistry
1 answer:
aliina [53]3 years ago
3 0

Answer:

B) Iron (c=0.45 J/g°C)

Explanation:

Given that:-

Heat gain by water = Heat lost by metal

Thus,  

m_{water}\times C_{water}\times (T_f-T_i)=-m_{metal}\times C_{metal}\times (T_f-T_i)

Where, negative sign signifies heat loss

Or,  

m_{water}\times C_{water}\times (T_f-T_i)=m_{metal}\times C_{metal}\times (T_i-T_f)

For water:

Mass = 120 g

Initial temperature = 21.8 °C

Final temperature = 24.5 °C

Specific heat of water = 4.184 J/g°C

For metal:

Mass = 40.2 g

Initial temperature = 99.3 °C

Final temperature = 24.5 °C

Specific heat of metal = ?

So,  

120\times 4.184\times (24.5-21.8)=40.2\times C_{metal}\times (99.3-24.5)

40.2C_{metal}\left(99.3-24.5\right)=120\times \:2.7\times \:4.184

40.2C_{metal}\left(99.3-24.5\right)=1355.616

C_{metal}=0.45\ J/g^0C

<u>This value corresponds to iron. Thus answer is B.</u>

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Answer:

6. 7870 kg/m³ (3 s.f.)

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= 64.2 ÷1000 kg

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= 8.16 cm³

= 8.16 ÷ 10⁶

= 8.16 \times 10^{ - 6} \:  kg

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=  \frac{0.151} {1.25 \times 10^{ - 5} }  \\  = 12600 \: kg/ {m}^{3}

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6 0
3 years ago
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