Question

A metal piece of mass 50 g at 27°c requires 2400 j of heat energy in order to raise it's temperature to 327°c.calculate the specific heat capacity of the metal.

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Answer 1

Answer:

480 J/(kg.K)

Explanation:

using h=mst

h - heat

m-mass(convert it into s.i)

s- specific heat

t- temp diff.

Answer 2

$\\\;\underbrace{\underline{\sf{Understanding\;the\;Concept}}}$

Here the concept of Specific Heat has been used. We know the relationship between Specific Heat and Heat of a substance. We are given the heat, change in temperature and mass of the substance. Firstly we shall use equation and then rearrange it to get our required form. Then after applying values in that, we can get our answer.

Let's do it !!

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★ Formula Used :-

$\\\;\boxed{\sf{\pink{Heat\;=\;\bf{Specific\:Heat\:\times\:Mass\:\times\:Temperature}}}}$

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★ Solution :-

Given,

» Mass of the metal piece = m = 50 g

» Initial Temperature = T₁ = 27° C

» Final Temperature = T₂ = 327° C

» Heat required initially = H = 2400 J

• Let c be the Specific Heat of the Metal.

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~ For the mass of metal in S.I. unit ::

• SI unit of mass = Kg

We know that, for calculation of Specific Heat, we need the mass of substance in Kg for precise answer. Though we can calculate answer by using the given quantity, but still converting them to SI form gives us precise answer.

Then, we know that

✒ 1 Kg = 1000 g

✒ 1 g = 1/1000 Kg

Then,

✒ 50 g = 50/1000 Kg

✒ 50 g = 1/20 Kg

✒ 50 g = 0.05 Kg

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~ For ∆T of the metal ::

∆T shows the temperature difference of the metal. This is given by subtracting Initial temperature from Final Temperature.

Then,

✒ Temperature Difference, ΔT = T₂ - T₁

✒ ∆ T = 327 ° C - 27 ° C

✒ ∆T = 300 ° C

This is also called as Change in Temperature.

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~ For the Specific Heat of metal ::

We know that,

$\\\;\sf{:\rightarrow\;\;H\;=\;\bf{Specific\:Heat\:\times\:Mass\:\times\:Temperature}}$

• Here Temperature shows the Temperature Difference.

Now by applying values, we get

$\\\;\sf{:\rightarrow\;\;H\;=\;\bf{c\:\times\:m\:\times\:\Delta\:T}}$

$\\\;\sf{:\Longrightarrow\;\;2400\;=\;\bf{c\:\times\:0.05\:\times\:300}}$

$\;\sf{:\Longrightarrow\;\;2400\;=\;\bf{c\:\times\:15}}$

$\\\;\sf{:\Longrightarrow\;\;c\;=\;\bf{\dfrac{2400}{15}}}$

$\\\;\sf{:\Longrightarrow\;\;c\;=\;\bf{160\;\:J\:Kg^{-1}\:^{\circ}C^{-1}}}$

This is the required answer.

$\\\;\underline{\boxed{\tt{Specific\;Heat\;\:of\;\:metal\;=\;\bf{\purple{160\;\:J\:Kg^{-1}\:^{\circ}C^{-1}}}}}}$

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★ More to know :-

• Answer in CGS form :

According to CGS form,

→ Mass = 50 g

Now using this to calculate Specific Heat, we get

$\\\;\tt{:\mapsto\;\;2400\;=\;c\:\times\:50\:\times\:300}$

$\\\;\tt{:\mapsto\;\;2400\;=\;c\:\times\:1500}$

$\\\;\tt{:\mapsto\;\;c\;=\;\dfrac{2400}{15000}}$

$\\\;\tt{:\mapsto\;\;c\;=\;\bf{0.16\;\:J\:g^{-1}\:^{\circ}C^{-1}}}$

This gives our answer in CGS unit form.

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• Other Formulas related to this section ::

$\\\;\sf{\leadsto\;\;Latent\;Heat\;=\;\dfrac{Heat}{Mass}}$

$\\\;\sf{\leadsto\;\;Entropy\;=\;\dfrac{Heat}{Temperature}}$

$\\\;\sf{\leadsto\;\;Boltzmann's\;Constant\;=\;\dfrac{Energy}{Temperature}}$

$\\\;\sf{\leadsto\;\;Universal\;Gas\;Constant\;=\;\dfrac{PV}{nT}}$

$\\\;\sf{\leadsto\;\;Solar\;Constant\;=\;\dfrac{Energy}{Area\;\times\;Time}}$