Question

1. A 48.0−g sample of an unknown metal at 99°C was placed in a constant-pressure calorimeter containing 70.0 g of water at 24.0°C. The final temperature of the system was found to be 28.4°C. Calculate the specific heat of the metal. (The heat capacity of the calorimeter is 10.4 J/°C.)

2.A balloon 19.0 m in diameter is inflated with helium at 23°C. Calculate the work done (in kJ) during the inflation process if the atmospheric pressure is 97.7 kPa.

3.Ice at 0.0°C is placed in a Styrofoam cup containing 360 g of a soft drink at 25.8°C. The specific heat of the drink is about the same as that of water. Some ice remains after the ice and soft drink reach an equilibrium temperature of 0.0°C. Determine the mass of ice that has melted. Ignore the heat capacity of the cup. (Hint: It takes 334 J to melt 1 g of ice at 0.0°C.)

4. When 1.049 g of naphthalene (C10H8) is burned in a constant-volume bomb calorimeter at 298 K, 42.16 kJ of heat is evolved.

Calculate ΔU and w for the reaction on a molar basis.

Answer 1

1. We know

                     heat lost by metal = heat gained by water + heat gained by calorimeter …(1)

[Heat (Q) = mass(m) * specific heat capacity(C_p )* Temperature difference (ΔT)]….(2)

Given,   specific heat capacity of water = C_p,water = 4.184J/gºC

Heat capacity of the calorimeter = 10.4 J/°C

m_metal = 48.0g

m_water = 70.0g

Intial Temperature of metal (T_i,metal)= 99.0°C

Intial Temperature of calorimeter/water (T_i) = 24.0°C

Final Temperature of calorimeter/water/metal (T_f) = 28.4°C

From equation (1),

                - m_metal*C_p,metal*(T_f-T_i,metal)= [m_water*C_p,water+C_{p,calorimeter}](T_f-T_i)

                 -48.0g* C_p,metal*(28.4°C - 99.0°C)= [70.0g*4.184J/gºC+10.4 J/°C]*(28.4°C -24.0°C)

• C_p,metal = (303.28 J/°C*4.4°C)/(48.0g*70.6°C)
•             =0.394 J/gºC

Hence specific heat of the metal is 0.394 J/gºC

2. Before inflation, the volume of the balloon is zero. [V_i = 0]

Let suppose balloon as a spherical shape.

Final volume (V_f) = 4/3 πr³  

Given, diameter of balloon (d) = 19.0 m

            Radius of balloon (r) = d/2 = 9.5 m

i.e. V_f = 4/3 π(9.5m)³   = 3589.54 m³

         P_ext = 97.7 kPa = 97.7*10³ Pa

Work done = -P_extΔV = -P_ext(V_f-V_i) = -(97.7*10³ Pa)( 3589.54 m³ -0)

                  = -350698.058*10³ Pa-m³ =-350698.058*10³ J

                  = -350698.058 kJ ~ -3.507*10⁵ kJ

{negative sign indicates the work is done by the system.}

Hence the work done is 3.507*10⁵ kJ

3. Here, all the heat lost by the soft-drink is gain by the ice. So,

                                    heat gained by ice = heat lost by soft-drink …..(1)

Q_ice = m_ice*∆H_fus
Q_soft-drink =m_soft-drink*C_p*ΔT

Putting these in (1)

                                     m_ice*∆H_fus = -m_soft-drink*C_p*ΔT [negative sign for heat loss]…(2)

We know,

specific heat capacity of water = C_p,water = 4.184J/gºC =specific heat capacity of soft-drink

Given,   ∆H_fus = 334 J [It takes 334 J to melt 1 g of ice at 0.0°C. This is known as heat of fussion]

T_i = 25.8°C

T_f = 0.0°C

m_soft-drink = 360g

Putting the above values in (2)

m_ice*334 J = -360g*4.184J/gºC*(0.0°C - 25.8°C)

                          =>                   m_ice = (360g*4.184J/gºC*25.8°C)/( 334 J)

                          =>                    m_ice = 116.35g

Hence, mass of ice that has melted is 116.35 g

4. According to the first law of thermodynamics

ΔU = q + w ……………(1)

Where, w= work done = -P_extΔV

Here, volume is constant. So, ΔV = 0

Hence, w = 0

Now, equation (1) become

ΔU = q

Given, heat evolved (q) = -42.16 kJ [negative sign indicates the heat evolved from the system]

ΔU = -42.16 kJ

On the molar basis,

Molar mass of naphthalene (C₁₀H₈) = 128 g/mol

Mass of naphthalene given = 1.049 g

ΔU = (-42.16 kJ*128 g/mol)/(1.049 g) = -5144.40 kJ/mol (negative sign indicates the heat lost from the system)