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Question: The specific heat of a substance at temperature \[{t^ \circ }C\] is \(s = a{t^2} + bt + c\). The amo...

The specific heat of a substance at temperature tC{t^ \circ }C is s=at2+bt+cs = a{t^2} + bt + c. The amount of heat required to raise the temperature of mm grams of a substance from 0C{0^ \circ }C to tC{t^ \circ }C.

Explanation

Solution

Specific heat capacity (s)\left( s \right) is the amount of heat required to raise the temperature of unit mass of the substance through 1C{1^ \circ }C or (1K)\left( {1K} \right) and heat capacity(C)\left( {{C^|}} \right) is the amount of heat required to raise the temperature of the whole body by 1C{1^ \circ }C.

Complete step by step answer:
Given: s=at2+bt+cs = a{t^2} + bt + c ............(1)............\left( 1 \right)
Mass=m = m and temperature of the substance changes from 0C{0^ \circ }C to tC{t^ \circ }C.
Let qq be the amount of heat required to raise the temperature of mm grams of substances from 0C{0^ \circ }C to tC{t^ \circ }C.
We know that, heat in terms of specific heat capacity can be given as
q=0tmsdtq = \int\limits_0^t {msdt} ............(2)............\left( 2 \right)
Amount of heat required q=0tmsdtq = \int\limits_0^t {msdt}
Substituting equation (1)\left( 1 \right) in equation(2)\left( 2 \right), we get
q=0tm(at2bt+c)dt\Rightarrow q = \int\limits_0^t {m\left( {a{t^2}bt + c} \right)} dt (\because Temperature of the substance changes from 0C{0^ \circ }C to tC{t^ \circ }C)
On integrating with respect to tt , the above equation becomes
q=m×(a3t3+b2t2+ct)0t\Rightarrow q = m \times \left( {\dfrac{a}{3}{t^3} + \dfrac{b}{2}{t^2} + ct} \right)_0^t (xndx=xn+1n+1)\left( {\because \int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}}} } \right)
On applying the limits, we get
q=m×[a3(t30)+b2(t20)+c(t0)]\Rightarrow q = m \times \left[ {\dfrac{a}{3}\left( {{t^3} - 0} \right) + \dfrac{b}{2}\left( {{t^2} - 0} \right) + c\left( {t - 0} \right)} \right]
On simplifying the above equation
q=m(at33+bt22+ct)\therefore q = m\left( {\dfrac{{a{t^3}}}{3} + \dfrac{{b{t^2}}}{2} + ct} \right)

Therefore, heat required to raise the temperature of mass m from 0C{0^ \circ }C to tC{t^ \circ }C is q=m(at33+bt22+ct)q = m\left( {\dfrac{{a{t^3}}}{3} + \dfrac{{b{t^2}}}{2} + ct} \right).

Note: Heat is a form of internal energy which is obtained due attractive force of molecules present in a body and random movement of molecules whereas temperature is the quantity which is necessary to determine the direction in which heat flows when two bodies are kept in contact. The S.IS.I unit of heat is joules(J)\left( J \right) and the S.IS.Iunit of temperature is Kelvin (K)\left( K \right).