Answer :
m₁ = mass of water = 75 g
T₁ = initial temperature of water = 23.1 °C
c₁ = specific heat of water = 4.186 J/g°C
m₂ = mass of limestone = 62.6 g
T₂ = initial temperature of limestone = ?
c₂ = specific heat of limestone = 0.921 J/g°C
T = equilibrium temperature = 51.9 °C
using conservation of heat
Heat lost by limestone = heat gained by water
m₂c₂(T₂ - T) = m₁c₁(T - T₁)
inserting the values
(62.6) (0.921) (T₂ - 51.9) = (75) (4.186) (51.9 - 23.1)
T₂ = 208.73 °C
in three significant figures
T₂ = 209 °C
Answer : The initial temperature of the limestone is [tex]1.05\times 10^{2}^oC[/tex].
Solution : Given,
Mass of limestone = 62.6 g
Mass of water = 75 g
Final temperature of limestone = [tex]51.9^oC[/tex]
Final temperature of water = [tex]51.9^oC[/tex]
Initial temperature of water = [tex]23.1^oC[/tex]
The specific heat capacity of limestone = [tex]0.921J/g^oC[/tex]
The specific heat capacity of water = [tex]4.186J/g^oC[/tex]
The formula used :
q = m × c × ΔT
where,
q = heat required
m = mass of an element
c = heat capacity
ΔT = change in temperature
According to this, the energy as heat lost by the system is equal to the gained by the surroundings.
Now the above formula converted and we get
[tex]q_{system}=-q_{surrounding}[/tex]
[tex]m_{system}\times c_{system}\times (T_{final}-T_{initial})_{system}= -[m_{surrounding}\times c_{surrounding}\times (T_{final}-T_{initial})_{surrounding}][/tex]
[tex]m_{limestone}\times c_{limestone}\times (T_{final}-T_{initial})_{limestone}= -[m_{water}\times c_{water}\times (T_{final}-T_{initial})_{water}][/tex]
Now put all the given values in this formula, we get
[tex]62.6g\times 0.921J/g^{0}C \times (51.9^{0}C-T_{\text{ initial limestone}})= -[75g\times 4.186J/g^{0}C \times (51.9^{0}C-23.1^oC)][/tex]
By rearranging the terms, we get the value of initial temperature of limestone.
[tex]T_{\text{ initial limestone}}=104.926^{0}C=1.05\times 10^{2}^oC[/tex]
Therefore, the initial temperature of the limestone is [tex]1.05\times 10^{2}^oC[/tex].