ssume all temperatures to be exact.
A 0.240-kg coffee cup at 27∘C is filled with 0.200kg of brewed coffee at 96∘C. The cup and the coffee come to thermal equilibrium at 84∘C.
Part A
If no heat is lost to the environment, what is the specific heat of the cup material? [Hint: Consider the coffee essentially to be water.]
the sum of the heats gained is zero.
Heat gained by cup+heatgained coffee=0
.240ccup*(84-27)+.200*cwater*(84-96)=0
put in the value for thespecifice heat of water (cwater), solve for ccup.
To determine the specific heat of the cup material, we can use the principle of conservation of energy. The heat gained by the coffee must be equal to the heat lost by the cup.
The equation for heat transfer is given by:
Q = mcΔT
Where:
Q = heat transfer (in Joules)
m = mass (in kilograms)
c = specific heat capacity (in Joules per kilogram per degree Celsius)
ΔT = change in temperature (in degrees Celsius)
We know the following values:
For the coffee:
m1 = 0.200 kg (mass of coffee)
T1 = 96°C (initial temperature of coffee)
Tf = 84°C (final temperature of coffee)
c1 = specific heat capacity of water = 4,186 J/kg·°C (since coffee is essentially water)
For the cup:
m2 = 0.240 kg (mass of cup)
T2 = 27°C (initial temperature of cup)
Tf = 84°C (final temperature of cup)
c2 = specific heat capacity of cup material (what we want to find)
Using the principle of energy conservation, we can equate the heat gained by the coffee to the heat lost by the cup:
Q1 = Q2
mc1ΔT1 = mc2ΔT2
Substituting the known values:
(0.200 kg)(4,186 J/kg·°C)(96°C - 84°C) = (0.240 kg)(c2)(84°C - 27°C)
Simplifying the equation:
(0.200 kg)(4,186 J/kg·°C)(12°C) = (0.240 kg)(c2)(57°C)
Solving for c2:
c2 = [(0.200 kg)(4,186 J/kg·°C)(12°C)] / [(0.240 kg)(57°C)]
c2 ≈ 878 J/kg·°C
Therefore, the specific heat of the cup material is approximately 878 J/kg·°C.