A piece of metal of mass 25 g at 94◦C is placed in a calorimeter containing 57.4 g of water at 24◦C. The final temperature of the mixture is 34.2 ◦C. What is the specific heat capacity of the metal?Assume that there is no energy lost to the surroundings. Answer in units of J/g·◦C.
See your other post. Apparently you ignore the Ccal.
To find the specific heat capacity of the metal, we can use the equation:
Q = mcΔT
Where:
Q = heat transferred
m = mass of the substance
c = specific heat capacity
ΔT = change in temperature
In this case, we know the mass of the metal (m) is 25 g, the initial temperature (Ti) is 94◦C, the final temperature (Tf) is 34.2◦C, and the specific heat capacity of water is 4.18 J/g·◦C.
First, let's calculate the heat transferred to the metal:
Qmetal = mcΔT
Qmetal = (25 g)(c)(34.2◦C - 94◦C)
Qmetal = -1500 c
Next, let's calculate the heat transferred to the water:
Qwater = mcΔT
Qwater = (57.4 g)(4.18 J/g·◦C)(34.2◦C - 24◦C)
Qwater = 2347.64 J
Since energy is conserved in this system, the heat transferred from the metal to the water is equal to the heat transferred from the water to the metal:
Qmetal = -Qwater
-1500 c = 2347.64 J
Now we can solve for c:
c = -2347.64 J / -1500 g
c ≈ 1.565 J/g·◦C (rounded to three decimal places)
Therefore, the specific heat capacity of the metal is approximately 1.565 J/g·◦C.