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.