Question

Test Object	Object Ti (°C)	Water Ti (°C)	Object + Water Tf (°C)	Water DT = Tf - Ti (°C)	Object DT = Tf - Ti (°C)	Object mass (g)	Water Mass (g)
Steel bolt	81 C	20 C	36 C	16 C	65 C	46 g	150 g

1. The heat lost by the hot bolt is equal to the heat gained by the water in the calorimeter. Use the equations provided in the eScience manual and what you know about heat to solve for the specific heat of the steel. Show your work.
2. What is the specific heat of steel from Table 1 in the eScience manual? Find the percent error for your calculated specific heat relative to this accepted value. Show your work.
3. What was the difference between your calculation and the actual specific heat of steel? How could you improve the experiment to reduce this error?

The equation for specific heat given: m1c1(change in temp. 1) = m2c2(change in temp.2)

Specific heat of steel from Table 1 is 0.46J/gC

Answer 1

(i) Heat lost by the hot bolt = Heat gained by the water in the calorimeter

$\Delta$Q_lost = $\Delta$Q_gained

m_s C_s'$\Delta$T = m_w C_w$\Delta$T

where, m_s = mass of steel bolt = 46 g

m_w = mass of water = 150 g

C_w = specific heat of water = 4.186 J/g ⁰C

C_s' = specific heat of steel bolt = ?

then, we get

(46 g) C_s' (65 ⁰C) = (150 g) (4.186 J/g ⁰C) (16 ⁰C)

(2990 g ⁰C) C_s' = (10046.4 J)

C_s' = [(10046.4 J) / (2990 g ⁰C)]

C_s' = 3.36 J/g ⁰C

(ii) From Table (1) in the eScience manual, the specific heat of steel which will be given as -

C_s = 0.46 J/g ⁰C

The percent error for our calculated specific heat relative to this accepted value which will be given by -

% Error = { |(Theoretical value) - (Experimental value)| / (Experimental value) } x 100

% Error = { |(0.46 J/g ⁰C) - (3.36 J/g ⁰C)| / (3.36 J/g ⁰C) } x 100

% Error = 86.3%

(iii) The difference between our calculation and the actual specific heat of steel which will be given as -

$\Delta$C_steel = | (actual value) - (calculated value) |

$\Delta$C_steel = |(0.46 J/g ⁰C) - (3.36 J/g ⁰C)|

$\Delta$C_steel = 2.9