Pondering the concept of specific heat while walking home from chemistry class,
ID: 556378 • Letter: P
Question
Pondering the concept of specific heat while walking home from chemistry class, you find an old penny on the ground and begin to wonder what the specific heat of copper is. You rush home and put 5.00 g of water at 18.00 °C in a styrofoam cup. You heat the 2.45 g penny to 50.00 °C and drop it in the water. The final temperature of the metal and water is 19.38 °C. Assuming the penny is pure copper (which modern pennies are not), what is the specific heat of copper? The specific heat of water is 4.18
Explanation / Answer
Copper:
Mass of copper penny m1 = 2.45 g
specific heat of coper c1 = ?
temperatute t1 = 50 oC
water:
mass of water m2 = 5 g
specific heat of water c2 = 4.18 J/g/oC
temperatute t2 = 18 oC
Given that final temperature of the metal and water t = 19.38 °C
Then,
Heat lost by copeer penny at 50 °C = Heat gained by water at 18 oC
Heat loss or gain Q = mcdT , c= specific heat
Hence,
m1.c1.(t1-t) = m2.c2.(t-t2) , t is the final temperature.
2.45 g x c1 x ( 50-19.38 ) = 5 g x 4.18 J/g/oC x (19.38 -18)
On simplification,
c1 = 0.384 J/g/oC
Therefore,
specific heat of copper = 0.384 J/g/oC