Question

If 100 g of aluminum at 90°C were placed in a calorimeter with 50 g of water 20 °C and the resulting temperature of the mixture was 28 °C, calculate the calorimeter constant for the calorimeter used in this experiment. (specific hea of water = 4.184 J/g °C and specific heat of aluminum = 0.900 J/g °C). O 390.6 J/g °C O 906.7 J/g °C O 488.3 J/g °C O 0.456 J/g °C O 336.7 J/g °C

Answer 1

Calorimeter constant is defined as the quantity of heat capacity of calorimeter.

it simple baised on the concept

Heat lost by hot body = heat gained by cold body

avel calculation Calorimeter constant Griven? weight of Al = 100 g for 90°c Temp of Al = weight of water Temp of wales- 50g & cold Cold 200 J cola final temp = 20°c specific heat capacity of water - " » Al = 4.18 4 J/ge 0.9005/9°c. AT for cold body (water) Trinid - Tualen - atrod 20 - 20 – de 7 at for Hot Tal body EAI) & Tinal - = 28-90 TDT Hor = - 6200 l # Heat lost by Hot body (AI) = m Cp At Temp change mass Heat capacily = 100 x 0.goox - 62 Pror = – 5500 5

Heat gained by cold body (water) - most Pecolop - 50x 4.184x 8 - 1613.657 We know - OR Scold & Ical = 0 - Hot = cold + I can mor & Go C alorimeter heat se car = – 2or - Gold = 5500 - 1613 6. Ical - +3906.4 T of calorimeter? – Calorimeten constant or Heat capacity Cecalorimeter AT cold 3906.4 Calorimeter constant = 488.3 goc