How much heat is required to raise the temperature of a 50.0 g block of iron by
#10.0^@"C"# ? Specific heat of iron: #"0.449 J/g"""^@"C"#

Question

Wyatt Adkins · Accepted Answer

#"225 J"#

The key to this problem is the specific heat of iron, which is said to be equal to

#c_"iron" = "0.449 J g"^(-1)""^@"C"^(-1)#

As you know, the specific heat of a substance tells you how much heat is needed to raise the temperature of a sample of #"1 g"# by #1^@"C"#.

Keep in mind that this represents the amount of heat needed per gram, per degree Celsius!

So, how much heat would be needed to increase the temperature of a #"50.0 g"# block of iron by #1^@"C"# ?

#50.0 color(red)(cancel(color(black)("g"))) * overbrace("0.449 J" /(1color(red)(cancel(color(black)("g"))) ""^@"C"))^(color(blue)("the specific heat of iron")) = "22.45 J" ""^@"C"^(-1)#

This means that in order to increase the temperature of #"50.0 g"# of iron by #1^@"C"# you need to provide it with #"22.45 J"# of heat.

You can thus say that in order to increase the temperature of the block by #10.0^@"C"#, you must provide ten times as much heat

#10.0 color(red)(cancel(color(black)(""^@"C"))) * overbrace("22.45 J"/(1color(red)(cancel(color(black)(""^@"C")))))^(color(blue)("for 50.0 g of iron")) = color(green)(bar(ul(|color(white)(a/a)color(black)("225 J")color(white)(a/a)|)))#

The answer is rounded to three sig figs.

Olivia Anton · Answer

undefined To calculate the heat required to raise the temperature of a substance, you use the formula:

Q = m * c * ΔT

where:
Q is the heat energy (in joules),
m is the mass of the substance (in grams),
c is the specific heat capacity of the substance (in J/g°C),
ΔT is the change in temperature (in °C).

Plugging in the given values:

m = 50.0 g
c = 0.449 J/g°C
ΔT = 10.0°C

Q = (50.0 g) * (0.449 J/g°C) * (10.0°C) = 224.5 J

So, the heat required to raise the temperature of a 50.0 g block of iron by 10.0°C is 224.5 joules.

How much heat is required to raise the temperature of a 50.0 g block of iron by #10.0^@"C"# ? Specific heat of iron: #"0.449 J/g"""^@"C"#