A piece of gold initially at a temperature of 25.1 °C absorbs 675 J of heat, raising its temperature to 57.4 °C. Assuming the specific heat of gold is 0.126 J/(g°C), what is the mass of the sample?

Answer 1

#"166 g"#

The key here is the specific heat of gold, which is said to be equal to

#c_"gold" = "0.126 J g"^(-1)""^@"C"^(-1)#
This tells you that in order to increase the temperature of #"1 g"# of gold by #1^@"C"#, you need to provide it with #"0.126 J"# of heat.
In your case, the temperature increases from #25.1^@"C"# to #57.4^@"C"#, which implies that it changes by
#57.4^@"C" - 25.1^@"C" = 32.3^@"C"#
Now, you can use the specific heat of gold to calculate how much energy would be needed to increase the temperature of gold by #32.3^@"C"#.
#32.3 color(red)(cancel(color(black)(""^@"C"))) * "0.126 J"/("1 g" * 1color(red)(cancel(color(black)(""^@"C")))) = "4.0698 J g"^(-1)#
This tells you that in order to increase the temperature of #"1 g"# of gold by #32.3^@"C"#, you need to provide it with #"4.0698 J"# of heat.
You can thus say that #"675 J"# of heat will increase the temperature of
#675 color(red)(cancel(color(black)("J"))) * "1 g"/(4.0698color(red)(cancel(color(black)("J")))) = "165.86 g"#
of gold by #32.3^@"C"#. Rounded to three sig figs, the answer will be
#color(darkgreen)(ul(color(black)("mass of gold = 166 g")))#
Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer 2

Using the formula for heat transfer:

Q = mcΔT

where: Q = heat transferred (in joules) m = mass of the sample (in grams) c = specific heat capacity of gold (in J/(g°C)) ΔT = change in temperature (in °C)

Rearranging the formula to solve for mass:

m = Q / (c * ΔT)

Plugging in the given values:

m = 675 J / (0.126 J/(g°C) * (57.4°C - 25.1°C))

m ≈ 25.96 grams

Therefore, the mass of the sample is approximately 25.96 grams.

Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

Not the question you need?

Drag image here or click to upload

Or press Ctrl + V to paste
Answer Background
HIX Tutor
Solve ANY homework problem with a smart AI
  • 98% accuracy study help
  • Covers math, physics, chemistry, biology, and more
  • Step-by-step, in-depth guides
  • Readily available 24/7