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).