Question

Trial 1 29.60 g Cu 95.9 25.9 23.09 g 21.7 Mass of Pennies Used Initial Temperature of Pennies Final Temperature of the System Mass of Water in Calorimeter Initial Temperature of Water in Calorimeter Specific Heat of Water q Gained by Water Specific Heat of Copper (Assuming No Heat Lost to 4.184 J/g'C Calorimeter.) Average q U) Absorbed by 127.4J Calorimeter from Part B. Specific Heat of Copper (Using the data from Part B) If the specific heat of Cu is 0.386 VEC, what is the % error from the 96 above row?

Answer 1

Calculation of heat gained by water

Mass of water, m_water =23.09 g

Intial temperature of water, T_i = 21.7 ^oC

Final temperature of water, T_f = 25.9 ^oC

Change in temperature of water, $\Delta$T = T_f - T_i = 25.9 ^oC - 21.7 ^oC = 4.2 ^oC

Specific heat of water, c_water = 4.184 J/g^oC

The heat gained by water can be calculated as below

q_{water =} c_water * m_water* $\Delta$T

=(4.184 J/g^oC)(23.09 g)( 4.2 ^oC)

= 405.76 J

Calculation of specific heat of copper when no heat is lost to calorimeter

If no heat is lost to the calorimeter, then

Heat gained by water = Heat lost by copper pennies

q_{water =} -q_copper

Therefore, q_copper = - 405.76 J

Now,

Mass of copper pennies, m_copper =29.60 g

Intial temperature of pennies, T_i = 95.9 ^oC

Final temperature of pennies, T_f = 25.9 ^oC

Change in temperature of pennies, $\Delta$T = T_f - T_i = 25.9 ^oC - 95.9 ^oC = - 70 ^oC

Using,

q_copper = c_copper * m_copper * $\Delta$T

c_copper = q_copper / ( m_copper * $\Delta$T )

Plugging the values in the above equation

c_copper = ( - 405.76 J) / {29.60 g * ( - 70 ^oC)}

=0.196 J/ g ^oC

Calculation of specific heat of copper when 127.4 J of heat is absorbed by the calorimeter

If 127.4 J of heat is absorbed by calorimeter, then actual heat lost heat lost by copper =

-[Heat gained by water + heat gained by calorimeter] -----------(negative sign is for heat lost)

=-(405.76 J+ 127.4 J)

=-533.16 J

Plugging this value in equation below, we get

c_copper = q_copper / ( m_copper * $\Delta$T )

= (-533.16 J) / {29.60 g * ( - 70 ^oC)}

= 0.257 J/ g ^oC

Calculation of percentage error

Percentage error = {(Theoretical value - experimental value) / theoretical value} *100 %

Theoretical value = 0.385  J/ g ^oC

Experimental value = 0.257 J/ g ^oC  

Percentage error =[{( 0.385  J/ g ^oC ) - (0.257 J/ g ^oC ) }/ (0.385  J/ g ^oC  )] *100 %

=33.25 %