# What is the specific heat of the unknown metal sample?

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An irregular lump of an unknown metal has a measured density of 2.97 g/mL. The metal is heated to a temperature of 173 °C and placed in a graduated cylinder filled with 25.0 mL of water at 25.0 °C. After the system has reached thermal equilibrium, the volume in the cylinder is read at 34.0 mL, and the temperature is recorded as 40.6 °C. What is the specific heat of the unknown metal sample? Assume no heat is lost to the surroundings.

An irregular lump of an unknown metal has a measured density of 2.97 g/mL. The metal is heated to a temperature of 173 °C and placed in a graduated cylinder filled with 25.0 mL of water at 25.0 °C. After the system has reached thermal equilibrium, the volume in the cylinder is read at 34.0 mL, and the temperature is recorded as 40.6 °C. What is the specific heat of the unknown metal sample? Assume no heat is lost to the surroundings.

##### 2 Answers

#### Explanation:

**Write down given information for both substances (mass, specific heat, change in temperature)**

*For Water* :

*For Metal* :

**Use the formula #Q=mcDeltaT# to find the change in energy in water**.

This (

**Rearrange the formula #Q=mcDeltaT# to find #c# of the metal**.

#C_m ~~ "0.457 J/g"^@ "C"#

The **displacement** of water gives the volume and thus mass of the metal:

#"34.0 - 25.0 mL" xx "2.97 g"/"mL" = "26.73 g"#

By **conservation of energy**,

#q_m + q_w = 0# where

#m# is metal and#w# is water.

Thus

#q_m = -q_w#

and

#m_mC_mDeltaT_m = -m_wC_wDeltaT_w# where

#C# is the constant-pressure specific heat capacity in#"J/g"^@ "C"# ,#m_i# the mass, and#DeltaT# the change in temperature in#""^@ "C"# .

They both reach the same **thermal equilibrium temperature**, so

#m_mC_m(T_f - T_m) = -m_wC_w(T_f - T_w)#

Therefore, assuming the specific heat capacity of water stays *constant* in the temperature range, and that its *density* is that at

#color(blue)(C_m) = -(m_wC_w(T_f - T_w))/(m_m(T_f - T_m))#

#= -("25.0 mL" xx "0.9919880 g"/"mL" cdot "4.184 J/g"^@ "C" cdot (40.6^@ "C" - 25.0^@ "C"))/("26.73 g" cdot (40.6^@ "C" - 173^@ "C"))#

#= color(blue)("0.457 J/g"^@ "C")#