The molar heat capacity of a metal at low temperature varies with the temperature according to the equation C = bθ + (a/H)θ^3 where a, b and H are constant. How much heat per mole is transfered during the process in which the temperature change from 0.01H to 0.02H ?


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Question Number 58216 by Umar last updated on 20/Apr/19

$The molar heat capacity of a metal at$ $low temperature varies with the$ $temperature according to the equation$ $C = b θ + \frac{a}{H} θ^{3}$ $where a, b and H are constant .$ $How much heat per mole is transfered$ $during the process in which the$ $temperature change from 0 . 01 H$ $to 0 . 02 H ?$
Commented by Umar last updated on 20/Apr/19

$please help!$
	Answered by tanmay last updated on 20/Apr/19
	$it is a question of chemistry... however can be solved... C_(molar) =(dQ/n)→dQ=heat ,n=nos of mole,C_(molar) =molar heat capacity.here n=1mole i have doubt of unit of temparatue change from0.01H to 0.02H (as per question) what is H... C_(molar=) =(dQ/n) dQ=1×C Q=∫_θ_i ^θ_f 1×(bθ+(a/H)θ^3 )dθ =∣((bθ^2 )/2)+(a/H)×(θ^4 /4)∣_θ_i ^θ_r =(b/2)(θ_f ^2 −θ_i ^2 )+(a/(4H))(θ_f ^4 −θ_i ^4 ) =(b/2){(0.02)^2 −(0.01)^2 }+(a/(4H)){(0.02)^4 −(0.01)}$
	$i t i s a q u e s t i o n o f c h e m i s t r y . . .$ $h o w e v e r c a n b e s o l v e d . . .$ $C_{m o l a r} = \frac{d Q}{n} \to d Q = h e a t, n = n o s o f m o l e, C_{m o l a r} = m o l a r h e a t$ $c a p a c i t y . h e r e n = 1 m o l e$ $i h a v e d o u b t o f u n i t o f t e m p a r a t u e$ $c h a n g e f r o m 0.01 H t o 0.02 H (a s p e r q u e s t i o n)$ $w h a t i s H . . .$ $C_{m o l a r =} = \frac{d Q}{n}$ $d Q = 1 \times C$ $Q = \int_{θ_{i}}^{θ_{f}} 1 \times (b θ + \frac{a}{H} θ^{3}) d θ$ $=∣ \frac{b θ^{2}}{2} + \frac{a}{H} \times \frac{θ^{4}}{4} ∣_{θ_{i}}^{θ_{r}}$ $= \frac{b}{2} (θ_{f}^{2} - θ_{i}^{2}) + \frac{a}{4 H} (θ_{f}^{4} - θ_{i}^{4})$ $= \frac{b}{2} {{(0.02)}^{2} - {(0.01)}^{2}} + \frac{a}{4 H} {{(0.02)}^{4} - (0.01)}$
	Commented by Umar last updated on 20/Apr/19

	$thank you so much$ $i doughted the unit of temp too$
	Commented by tanmay last updated on 20/Apr/19

	$m o s t w e l c o m e . . .$
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