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

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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.

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

#### 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 offcarries 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 themassof the sample#c# is thespecific heatof the substance [equal to#"4.184 J g"^(-1)""^@"C"^(-1)# for liquid water]#DeltaT# is thechange 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

#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

#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 **do not** forget that you need to take

#-(- 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

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**.