Question

The specific heat of many solids at low temperature varies with absolute temperature t according to the relation s=at^3 where a is a constant. the heat energy required to raise the temperature of a mass m of such a solid from 0 to 20 is:

Answer 1

Answer:

Explanation:

dQ=msdt

integration(dQ) = m integration(sdt)

(limit 0 to 20)

Q=m integration(at^3dt)

(limit 0to 20)

Q=40000ma

Answer 2

Answer:

The heat energy required is $4*10^4ma$.

Explanation:

Given:

$s=at^3$

$t_1=0$

$t_2=20$

To find:

The heat energy required to raise the temperature (ΔQ)

Formula:

ΔQ = $\int\limits^{t_2}_{t_1} {ms } \, dt$

Solution:

Substituting the given values, we get

ΔQ = $\int\limits^{t_2}_{t_1} {ms } \, dt$

ΔQ = $\int\limits^{20}_{0} {mat^3} \, dt$

Since a and m are constants, we take them out of the integration. Therefore,

ΔQ = $ma\int\limits^{20}_0 {t^3} \, dt$

ΔQ = $ma[\frac{t^4}{4} ]^{20}_0$

ΔQ = $ma[\frac{20^4}{4} - 0]$

ΔQ = $ma[\frac{20^4}{4} ]$

ΔQ = 40000ma

ΔQ = 4×10^4 ma

Therefore, the heat energy required to raise the temperature of a mass m of a given solid from 0 to 20 is 4×10^4 ma.

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