Question

In the laboratory, a student uses a coffee cup calorimeter to determine the specific heat of a metal. She heats 18.1 grams of lead to 99.50°C and then drops it into a cup containing 81.7 grams of water at 23.00°C. She measures the final temperature to be 23.67°C. Assuming that all of the heat is transferred to the water, she calculates the specific heat of lead to be _____ J/g°C.

Answer 1

Answer:

$specificheat=0.177J/g^{0}C$

Explanation:

The heat evolved by lead will be absorbed by water.

The heat evolved by lead will be:

$Q1=massXspecificheatXchangeintemperature$

$Q1=18.1X(Specificheat)X(99.50-23.67)$

Heat absorbed by water will be:

$Q2=massXspecificheatX(changeintemperature)$

$Q2=81.7X4.184(23.67-23)=229.03J$

We know that

Q1=Q2

$18.1X(Specificheat)X(99.50-23.67)=229.03$

$1372.523Xspecificheat=229.03$

$specificheat=0.177J/g^{0}C$