A student wants to cool 250g of diet 7-up initially at 27 degree celcious by adding ice initially at -18 degree C. How much ice should he add so that the final temperature will be 0 degree C with all the ice melted?[Assume that the container is not taking any heat and specific heat of 7 up 4186 J/K.degree Celcious.specific heat of ice =2000 J/K degree Celcious and the latent heat of ice =3.34*105]

Question

Accepted Answer

ঘ. 76.4g

ব্যাখ্যা: ডায়েট সেভেন আপ কর্তৃক বর্জিত তাপ $Q_1 = m_1 s_1 \Delta 	heta_1 = 0.250 	imes 4186 	imes (27 - 0) = 28255.5	ext{ J}$। বরফ কর্তৃক গৃহীত তাপ $Q_2 = m_2 [s_{	ext{ice}} (0 - (-18)) + L_f] = m_2 [2000 	imes 18 + 3.34 	imes 10^5] = m_2 (36000 + 334000) = 370000 m_2$। তাপীয় সাম্যাবস্থায় $Q_1 = Q_2 \implies m_2 = \frac{28255.5}{370000} \approx 0.07637	ext{ kg} = 76.4	ext{ g}$।

পদার্থবিজ্ঞান প্রশ্ন

সম্পর্কিত প্রশ্ন