Question #836a3

3 Answers
Apr 11, 2015

The specific heat capacity of the metal is 3.5 J°C⁻¹g⁻¹.

There are two heats involved:

Heat absorbed by metal + heat lost by water = 0

#q_1 + q_2 = 0#

#m_1c_1ΔT_1 + m_2c_2ΔT_2 = 0#

The density of water is 1.00 g/mL. So

#m_2 = 20.0 cancel("mL water") × "1.00 g water"/(1 cancel("mL water")) = "20.0 g water"#

#m_1 = "5.10 g"#; #c_1 = "?"#; #ΔT_1 = ("48.6 – 26.8) °C" = "21.8 °C"#
#m_2 = "20.0 g"#; #c_2 = "4.184 J°C⁻¹g⁻¹"#; #ΔT_2 = "(22.1 – 26.8) °C" = "-4.7 °C"#

#q_1 + q_2 = 0#

#"5.10 g" × c_1 × "21.8 °C" + 20.0 cancel("g") × "4.184 J·"cancel("°C⁻¹g⁻¹") × (-4.7 cancel("°C")) = 0#

#111.2 c_1 " g°C " – "393 J" = 0#

#c_1 = "393 J"/"111.2 g°C" = "3.5 J·g"^-1°"C"^-1# (2 significant figures)

Note: The answer can have only 2 significant figures, because #ΔT_2# has only 2 significant figures.

It looks as if your metal is lithium.

Don't do this experiment, because lithium reacts violently with water.

c2.staticflickr.com

Apr 11, 2015

Answer:

#c_"metal" = 3.53"J"/("g" ^@"C")#

Explanation:

So, you've got an unknown metal that weighs 5.10 g and has a temperature of 48.6 degrees Celsius.. You place it in a calorimeter that contains 20.0 mL of water initially at 22.1 degrees Celsius.

Notice that the temperature of the water increases and the temperature of the metal decreases.

You know that the heat lost by the metal was gained by the water. Mathematically, this is written as

#-q_"metal" = q_"water"# #-># the minus sign designates heat lost.

This equation becomes

#m_"metal" * c_"metal" * DeltaT_"metal" = m_"water" * c_"water" * DeltaT_"water"#

Notice that you were given a volume of water, not a mass, which implies you must use water's density at #22.1^@"C"#, which is listed at #"0.99775 g/mL"#, to determine its mass.

#rho = m/V => m = rho * V#

#m = "0.99775 g"/cancel("mL") * "20.0"cancel("mL") = "19.96 g"#

So, plug all your data into the main equation and solve for #c_"metal"#

#-["5.10 g" * c_"metal" * (26.8 - 48.6)^@"C"] = 19.96cancel("g") * 4.18"J"/(cancel("g") * ^@cancel("C")) * (26.8 - 22.1)^@cancel("C")#

#111.18 * c_"metal" = 392.036#

#c_"metal" = "392.036"/"111.18" = color(green)(3.53"J"/("g" * ^@"C"))#

Apr 12, 2015

You use the equation #Q_m=Q_w#
Where the #Q#'s are the heat exchanged
(and the suffixes are for metal and water respectively)

In general #Q=c*m*DeltaT#
#c=#specific heat
#m=#mass (in #g#)
#DeltaT=#change in temperature

Let's set up the equation and then fill in what we know:
#Q_m=Q_w#
#c_m*m_m*DeltaT_m=c_w*m_w*DeltaT_w#
#c_m*5.10*(48.6-26.8)=4.184*20.0*(26.8-22.1)#
You now have an equation with only one unknown, and this you can work out, can't you?

Note:
The #4.184# is the specific heat of water, which can be found in any book or site.