Question

when 3.425kj of heat ia added to calorimeter 55.0g of water temp rises from 28 Celsius to 39.54 Celsius the specific heat of water is 4.184g

U. Wully Ulica al 15 IUSE UY 21 dl l dlur UMUR GRILLEN - C. quantity of heat gained by 1.0 g of aluminum during an energy transfer process D. quantity of heat that changes the temperature of 27 g of aluminum by 1°C 10. When 3.425 kJ of heat is added to a calorimeter containing 55.00 g of water the temperature rises from 28.00°C to 39.54°C. Calculate the heat capacity of the calorimeter in J/S (The specific heat of water is 4.184 J/g °C) A. 66.7 J/°C B.63.9 JOC C. 27.3 JOSE D. 273 JOC 11. Calculate the energy (in kcal) lost when 37.5 g of gold cools from 224°C to 110°C. (specific heat capacity of gold = 0.13 J/g-K) A. 55.6 kcal B. 555.8 kcal C. 0.133 kcal D. 0.488 kcal 12. Which of the following does not take the shape of its container and has a definite volume? A. aluminum powder B. carbon dioxide (CO2) gas C mercury metal D. copper wire

Answer 1

Solution Total heat added = 3.425 kJ Heat required (2) to water to increase its temperature by 39.54 from 28,000 € 2= mx sx ot m = mass of H₂O = 550g S= specific beat of H20 - 4,184 I/go at = (89.54 - 28.00) = 11:54°C change in temperature in FFAASSLPDDDDDSDOMMAGADDDDDFTTPD So, Q = 55.0 x 4.184 X11.54 2= 2.655 KT Hence remaining heat = Total heat added - Heat taken by H₂O = 3.425 - 2,655 = 0.77 kJ This heat will be supplies to Calorymeter.

Heat Capacity = of colorimeter Heat supplies to colorimeter change in temperature 0.77 kJ 39.54- 28.00 0.77 11.54 = 0.0667 kJ/°C = 66,75/00 option (A) is correct.