A piece of copper block of mass 24g at 230 degree celsius is placed in a copper calorimeter of mass 60g containing 54g of water at 31 degree celsius.Assuming heat losses are negligible,calculate the final steady temperature of the mixture
(Specific heat capacity of water=4200J/kg/k)(Specific heat capacity /of copper=400J/kg/k)
Heat lost= 0.024×400(230-T)
Heat gained by water= 0.05×4200(T-31)
Heat gained by calorimeter=0.060×400(T-31)
Heat lost= Heat gained
= 0.024×400(230-T) = 0.54×4200
(T-31)+0.60×400(T-31)
T= 38. 34°C
to get the answer
Yes
Faisat
To calculate the final steady temperature of the mixture, we need to first calculate the heat gained by the water and the copper separately, and then equate it to the heat lost by the water and the copper.
Let's start by calculating the heat gained by the water.
Heat gained by water (Qw) = mass of water × specific heat capacity of water × change in temperature
Given:
mass of water (mw) = 54g = 0.054kg
specific heat capacity of water (Cw) = 4200J/kg/K
change in temperature of the water (ΔTw) = final temperature - initial temperature = Tf - 31°C
Since the copper block and the copper calorimeter are in thermal equilibrium, their final temperature will be the same. Let the final temperature be Tf.
Next, let's calculate the heat gained by the copper block.
Heat gained by copper (Qc) = mass of copper × specific heat capacity of copper × change in temperature
Given:
mass of copper (mc) = 24g = 0.024kg
specific heat capacity of copper (Cc) = 400J/kg/K
change in temperature of the copper (ΔTc) = final temperature - initial temperature = Tf - 230°C
According to the principle of conservation of energy, the total heat gained is equal to the total heat lost.
Qw + Qc = heat lost by water + heat lost by copper
Since the heat losses are negligible, the heat lost by the water and the copper is negligible. Therefore, we can equate the heat gained by the water and the copper.
Qw + Qc = 0
Substituting the formulas for Qw and Qc:
(mw × Cw × ΔTw) + (mc × Cc × ΔTc) = 0
Now substitute the given values and solve for Tf:
(0.054kg × 4200J/kg/K × (Tf - 31°C)) + (0.024kg × 400J/kg/K × (Tf - 230°C)) = 0
Simplify the equation and solve for Tf using algebraic manipulations:
(226.8(Tf - 31)) + (9.6(Tf - 230)) = 0
226.8Tf - 7034.8 + 9.6Tf - 2208 = 0
236.4Tf - 9242.8 = 0
236.4Tf = 9242.8
Tf = 9242.8 / 236.4
Tf ≈ 39.12°C
Therefore, the final steady temperature of the mixture is approximately 39.12°C.