Question

UL. HAL IVCI U IS DIUROFIN S 2. (15 pts) A 83.5 g sample of a metal alloy is heated to 88.1°C and it is then placed in a coffee-cup calorimeter containing 30.0 g water at 15.0°C. The final temperature of the metal + water is 25.3 "C. Calculate the specific heat of metal alloy, in J/g °C), assuming no heat escapes to the surroundings or is transferred to the calorimeter. The specific heat of water is 4.184 J/g °C). 3.(15pts) In each case, find the missing AH.(kJ/mole)

Answer 1

Answer 2 -

Given,

mass of Alloy = 83.5 g

Temperature of Alloy = 88.1^$\degree$ C

mass of water = 30.0 g

Temperature of water = 15.0^$\degree$ C

Final Temperature of Alloy = Final Temperature of water = Final Temperature of alloy + water (a+w) = 25.3^$\degree$ C

Specific Heat of water = 4.184 J/(g^$\degree$C)

Specific Heat of Alloy = ?

No heat escaped to surroundings or calorimeter

We know that,

Heat = mc$\Delta$T

where,

m = mass

c = Specific Heat

$\Delta$T = Change in Temperature (Final - Initial)

We know that heat lost by alloy is absorbed by water. So,

Heat of Alloy = - Heat of Water

mass of alloy * specific heat of Alloy (Final T_a- Initial T_a)=- mass of water * Specific Heat of Water (Final T_w - Initial T_w

specific heat of Alloy =- mass of water * Specific Heat of Water *(Final T_a+w - Initial T_w) / [mass of alloy * (Final T_a+w- Initial T_a)]

Put the values,

specific heat of Alloy = -30.0 g * 4.184 J/(g^$\degree$C) *(25.3^$\degree$ C - 15.0^$\degree$ C) / [83.5 g * (25.3^$\degree$ C- 88.1^$\degree$ C)]

specific heat of Alloy = -1292.856 J / -5243.8 g^$\degree$ C

specific heat of Alloy = 0.25 J/g^$\degree$ C [Answer]