How much heat in calories is needed to raise the temperature of "125.0 g" of Lead from 17.5^@"C" to 41.1^@"C"? (c_"Pb"=("0.130 J")/("g"*""^@"C"))

1 Answer
May 13, 2018

"95.6 cal" are needed.

Explanation:

Use the following equation:

q=mcDeltaT,

where:

q is heat energy, m is mass, c is specific heat capacity, and DeltaT is the change in temperature. DeltaT=T_"final"-T_"initial"

Known

m="125 g"

c_"Pb"=(0.130"J")/("g"*""^@"C")

T_"initial"="17.5"^@"C"

T_"final"="42.1"^@"C"

DeltaT="42.1"^@"C"-"17.5"^@"C"="24.6"^@"C"

Unknown

q

Solution

Plug the known values into the equation and solve.

q=(125"g")xx((0.130"J")/("g"*""^@"C"))xx(24.6^@"C")="400. J"
(rounded to three significant figures)

Convert Joules to calories

"1 J"="0.2389 cal" to four significant figures.

400. color(red)cancel(color(black)("J"))xx(0.2389"cal")/(1color(red)cancel(color(black)("J")))="95.6 cal"
(rounded to three significant figures)

"95.6 cal" are needed.