The specific heat of copper metal was determined by putting a piece of the metal weighing 33.6 g in hot water. The quantity of heat absorbed by the metal was calculated to be 47 J from the temperature drop of the water. What was the specific heat of the metal if the temperature of the metal rose 3.63°C?
The sum of the heats gained is zero (some will lose heat and be negative).
Heatgainedcopper+heatgained water=0
33.6g*c*3.63-47=0
solve for c
To determine the specific heat of copper, we can use the formula:
Q = mcΔT
Where:
Q is the quantity of heat absorbed by the metal (in joules)
m is the mass of the metal (in grams)
c is the specific heat of the metal (in J/g·°C)
ΔT is the change in temperature of the metal (in °C)
In this problem, we are given:
Q = 47 J
m = 33.6 g
ΔT = 3.63 °C
Rearranging the formula, we can solve for c:
c = Q / (mΔT)
Plugging in the given values:
c = 47 J / (33.6 g × 3.63 °C)
To calculate this, we need to convert the mass from grams to kilograms, and the change in temperature from Celsius to Kelvin.
First, we convert the mass:
m = 33.6 g = 0.0336 kg
Then, we convert the temperature:
ΔT = 3.63 °C = 3.63 K
Plugging in the converted values:
c = 47 J / (0.0336 kg × 3.63 K)
Now, we can calculate:
c ≈ 391 J/(kg·K)
Therefore, the specific heat of copper is approximately 391 J/(kg·K).