Question

A 62.6-gram piece of heated limestone is placed into 75.0 grams of water at 23.1°C. The limestone and the water come to a final temperature of 51.9°C. The specific heat capacity of water is 4.186 joules/gram degree Celsius, and the specific heat capacity of limestone is 0.921 joules/gram degree Celsius. What was the initial temperature of the limestone? Express your answer to three significant figures.

Answer 1

Answer : The initial temperature of the limestone is $1.05\times 10^{2}^oC$.

Solution : Given,

Mass of limestone = 62.6 g

Mass of water = 75 g

Final temperature of limestone = $51.9^oC$

Final temperature of water = $51.9^oC$

Initial temperature of water = $23.1^oC$

The specific heat capacity of limestone = $0.921J/g^oC$

The specific heat capacity of water = $4.186J/g^oC$

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

$q_{system}=-q_{surrounding}$

$m_{system}\times c_{system}\times (T_{final}-T_{initial})_{system}= -[m_{surrounding}\times c_{surrounding}\times (T_{final}-T_{initial})_{surrounding}]$

$m_{limestone}\times c_{limestone}\times (T_{final}-T_{initial})_{limestone}= -[m_{water}\times c_{water}\times (T_{final}-T_{initial})_{water}]$

Now put all the given values in this formula, we get

$62.6g\times 0.921J/g^{0}C \times (51.9^{0}C-T_{\text{ initial limestone}})= -[75g\times 4.186J/g^{0}C \times (51.9^{0}C-23.1^oC)]$

By rearranging the terms, we get  the value of initial temperature of limestone.

$T_{\text{ initial limestone}}=104.926^{0}C=1.05\times 10^{2}^oC$

Therefore, the initial temperature of the limestone is $1.05\times 10^{2}^oC$.

Answer 2

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