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An aluminum kettle weighs 1.05 $\mathrm{kg}$(a) What is the heat capacity of the kettle?(b) How much heat is required to increase the temperature of this kettle from $23.0^{\circ} \mathrm{C}$ to $99.0^{\circ} \mathrm{C} ?$(c) How much heat is required to heat this kettle from $23.0^{\circ} \mathrm{C}$ if it contains 1.25 $\mathrm{L}$ of water (density of 0.997 $\mathrm{g} / \mathrm{mL}$ and a specific heat of 4.184 $\mathrm{J} / \mathrm{g}^{\circ} \mathrm{C}$ )?
a$C_{\mathrm{Al}}=1345.5 \frac{\mathrm{J}}{\mathrm{c}}$b$q=71580.6 \mathrm{J}$c$q=467868.2 \mathrm{J}$
02:03
Aadit S.
Chemistry 101
Chapter 5
Thermochemistry
Drexel University
University of Kentucky
University of Toronto
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We're told that aluminum can always 1.05 kg. For a what is the heat capacity of the kettle? The capacity of the kettle sees equal the Mt capacity of the aluminum Mass is 1.05 kg. Karubas two g. Yeah, The capacity of aluminum is 897 jules per gram degrees Celsius. Therefore, the heat capacity of the kettle works out to 942 jewels per degrees Celsius. For B. How much heat is required to increase the temperature of the kettle just made of aluminum Uh from 23 to 99°C.. So the mass of the cattle, he is 1.05 kg. Which were these two g. Mhm. Just the capacity of aluminum is point 897 jewels per gram degrees C. And the temperature Is 9090C and it's initially at 23°C.. This will work out to 71 1000 580 .6 Jewels. For 71 6 kilo jewels. Yeah. Okay. For c How much heat is required to heat the kettle from 23°C.. I if it contains 1.25 L of water, don't feel well. First calculate the mass of the H 20 Which would equal the density times the volume. The massive or the density is .997 grams per mil. Leader. In terms of volume 1.25 L, convert this to milliliters And the mass of the water works out to 1246.25g, calculate the heat. Uh the water. M. C. Delta T. Mass of the waters. 1246.25 grams he capacity of water is 4.184 jewels per gram degrees Celsius, Temperatures going from 99°C.. initially at 23°C.. And the energy here works out to 396,000, 287 56 jewels, or 3.3396 your jewels. Therefore the heat required to heat the kettle and the water From 23 to 99°C would be the heat of the aluminum, plus the heat of the water Is equal to 71.6 killer jewels plus 396 killer jewels. So the total heat required here would be 468 killer jewels.
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