A cube of gold is heated then submerged in water. What is the volume of the cube of gold (in mL)?

A cube of gold is heated to a temperature of 94.2 °C, and then submerged in 31.3 mL (31.3 g) of H2O at an initial temperature of 28.7 °C. If the final temperature of the water is 45.6 °C, calculate the volume of the cube of gold (in mL). Gold is a metal with a specific heat of 0.130 J/(g •°C) and a density of 19.3 g/mL.

1 Answer
Feb 11, 2018

Answer:

#"18.1 mL"#

Explanation:

The idea here is that you must assume that the heat given off by the cube as it cools will be equal to the heat absorbed by the water as it warms.

#color(blue)(ul(color(black)(-q_"cube" = q_"water")))" " " "color(darkorange)("(*)")#

The minus sign is used here because, by definition, heat given off carries a negative sign.

Now, your tool of choice here will be the equation

#color(blue)(ul(color(black)(q = m * c * DeltaT)))#

Here

  • #q# is the heat given off/absorbed
  • #m# is the mass of the sample
  • #c# is the specific heat of the substance [equal to #"4.184 J g"^(-1)""^@"C"^(-1)# for liquid water]
  • #DeltaT# is the change in temperature, calculated as the difference between the final temperature and the initial temperature of the sample

You know that the final temperature of the gold + water system is equal to #45.6^@"C"#. This means that the change in temperature for the gold cube is equal to

#DeltaT_"cube" = 45.6^@"C" - 94.2^@"C" = - 48.6^@"C"#

Similarly, the change in temperature for the water is equal to

#DeltaT_"water" = 45.6^@"C" - 28.7^@"C" = 16.9^@"C"#

This means that if you take #m_"cube"# #"g"# to be the mass of the gold cube, you can say that the heat given off by the cube will be equal to

#q_"cube" = m_"cube" color(red)(cancel(color(black)("g"))) * "0.130 J" color(red)(cancel(color(black)("g"^(-1)))) color(red)(cancel(color(black)(""^@"C"^(-1)))) * (-48.6)color(red)(cancel(color(black)(""^@"C")))#

#q_"cube" = -(m_"cube" * 6.318) quad "J"#

The heat absorbed by the water will be equal to

#q_"water" = 31.3 color(red)(cancel(color(black)("g"))) * "4.184 J" color(red)(cancel(color(black)("g"^(-1)))) color(red)(cancel(color(black)(""^@"C"^(-1)))) * 16.9 color(red)(cancel(color(black)(""^@"C")))#

#q_"water" = "2,213.21 J"#

Next, use equation #color(darkorange)("(*)")# to find the mass of the gold cube--do not forget that you need to take #-q_"cube"# !

#-(- m_"cube" * 6.318) color(red)(cancel(color(black)("J"))) = "2,213.21" color(red)(cancel(color(black)("J")))#

You will end up with

#m_"cube" = "2,213.21"/6.318 = 350.3#

Since we've said that #m_"cube"# #"g"# represents the mass of the cube, you can say that the gold cube has a mass of #"350.3 g"#.

Now, you know the density of gold, so you can use its mass to find the volume of the cube.

#350.3 color(red)(cancel(color(black)("g"))) * "1 mL"/(19.3 color(red)(cancel(color(black)("g")))) = color(darkgreen)(ul(color(black)("18.1 mL")))#

The answer is rounded to three sig figs.