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3 years ago

If a 30 gram sample of Copper shots (specific heat = 0.385 J/gºC) changed from 27ºC to 90ºC, how much heat was involved?

Chemistry

2 answers:

Readme [11.4K]3 years ago

4 0

Answer:

A = 728J

Explanation:

Mass of substance = 30g

Specific capacity of copper = 0.385J/g°C

Initial temperature (T1) = 27°C

Final Temperature (T2) = 90°C

Heat energy (Q) = ?

Heat Energy (Q) = Mc∇T

Q = heat energy

M = mass of the substance

C = specific heat capacity of the substance

∇T = change in temperature of the substance = T2 - T1

Q = mc∇T

Q = mc×(T2 - T1)

Q = 30 × 0.385 × (90 - 27)

Q = 11.55 × 63

Q = 727.65J

The heat energy required to raise 30g of copper sample from 27°C to 90° is 728J

Natasha_Volkova [10]3 years ago

3 0

Answer:

728J

Explanation:

The following data were obtained from the question:

Mass (M) = 30g.

specific heat capacity (C) = 0.385 J/gºC.

Initial temperature (T1) = 27ºC

Final temperature (T2) = 90ºC

Change in temperature (ΔT) = T2 – T1 = 90°C – 27 = 63°C

Heat (Q) =...?

The heat involved can be obtained as illustrated below:

Q = MCΔT

Q = 30 x 0.385 x 63

Q = 728J

Therefore the heat involved is 728J

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Answer:

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Suppose that 0.48 g of water at 25∘C∘C condenses on the surface of a 55-gg block of aluminum that is initially at 25∘C∘C. If the

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Answer : The final temperature of the metal block is, $25^oC$

Explanation :

$heat_{absorbed}=heat_{released}$

As we know that,

$Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})$

$m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]$ .................(1)

where,

q = heat absorbed or released

$m_1$ = mass of aluminum = 55 g

$m_2$ = mass of water = 0.48 g

$T_{final}$ = final temperature = ?

$T_1$ = temperature of aluminum = $25^oC$

$T_2$ = temperature of water = $25^oC$

$c_1$ = specific heat of aluminum = $0.900J/g^oC$

$c_2$ = specific heat of water= $4.184J/g^oC$

Now put all the given values in equation (1), we get

$55g\times 0.900J/g^oC\times (T_{final}-25)^oC=-[0.48g\times 4.184J/g^oC\times (T_{final}-25)^oC]$

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We know that, 1 US dollar = 100 cents

1 cent = 1 US penny

So, 1 US dollar = 100 US pennies

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Let's make the set up starting with 1 penny as:

$1penny(\frac{$1}{100pennies})(\frac{1\mu g}{$1000})(\frac{1g}{10^6\mu g})(\frac{6.02*10^2^3atoms}{251g})$

= $2.40*10^1^0atoms$

Therefore, we can bye $2.40*10^1^0atoms$ of Cf in one US penny.

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tamaranim1 [39]

Answer:

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Explanation:

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From the balanced equation above,

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From the balanced equation above,

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Explanation:

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