Question

In the laboratory a student uses a "coffee cup" calorimeter to determine the specific heat of a metal. She heats 18.3 grams of gold to 99.38°C and then drops it into a cup containing 80.2 grams of water at 20.87°C. She measures the final temperature to be 21.40°C. Assuming that all of the heat is transferred to the water, she calculates the specific heat of gold to be J/g°C.

Answer 1

GIven : 1) mass of metal = 18.3 gm

2) T initail of metal = 99.38 ⁰C

3) T final of metal = 21.40 ⁰C

4) mass of water = 80.2 gm

5) T initail of water = 20.87 ⁰C

6) T final of water = 21.40 ⁰C

IN this case , heat is transfered from gold metal to water. Heat is lost by metal and absorbed by water. Hence we can write

q _gold = -q _water

   (C X m X T) gold = -   (C X m X ) water

where C ----> Specific capacity

m-----> mass

T-----> temp. difference.

C X 18.3 gm X ( 21.40-99.38) = -( 4.184 J/g^oC X 80.2 gm X ( 21.40-20.87))

C gold = 0.1246 J/ g^oC= Specific capacity of gold