When 72 g of a metal a t 97.0°C is added to 100 g of water at 25.0°C the final temperature is found to be 29.1°C. What is the heat capacity per gram of the metal? (Heat capacity of H2O
To find the heat capacity per gram of the metal, we can use the equation:
q = m × c × ΔT
Where:
q = heat transferred (in Joules)
m = mass (in grams)
c = specific heat capacity (in J/g°C)
ΔT = change in temperature (in °C)
Here, we have the mass of the metal (72 g) and the change in temperature (97.0°C - 29.1°C = 67.9°C). We also know the mass of water (100 g) and the change in temperature (29.1°C - 25.0°C = 4.1°C).
We need two equations to solve for the specific heat capacity of the metal. The first equation is for the heat transferred from the metal to the water:
q1 = m1 × c1 × ΔT1
The second equation is for the heat transferred from the water to the metal:
q2 = m2 × c2 × ΔT2
Since energy is conserved, we can set the two heat transfer equations equal to each other:
q1 = q2
m1 × c1 × ΔT1 = m2 × c2 × ΔT2
Substituting the given values:
(72 g) × c × (97.0°C - 29.1°C) = (100 g) × (4.18 J/g°C) × (29.1°C - 25.0°C)
Simplifying the equation:
72 × c × 67.9 = 100 × 4.18 × 4.1
Dividing both sides by 72 × 67.9:
c = (100 × 4.18 × 4.1) / (72 × 67.9)
Calculating this expression, we find:
c ≈ 0.345 J/g°C
Therefore, the heat capacity per gram of the metal is approximately 0.345 J/g°C.