A 12.9-g sample of an unknown metal at 26.5° is placed in a Styrofoam cup containing 50 g of water at 88.6°. The water cools down, and the metal warms up until thermal equilibrium is achieved at 87.1°. Assuming all the heat released by the water is absorbed by the metal and that the cup is perfectly insulated, determine the specific heat of the unknown metal. The specific heat of water is4,186 J/(kg ·K).
To determine the specific heat of the unknown metal, we can use the equation:
q = m * c * ΔT
Where:
q = heat gained or lost by the substance (in this case, the metal)
m = mass of the substance (in this case, the metal)
c = specific heat of the substance (what we need to find)
ΔT = change in temperature
First, we need to calculate the heat gained by the water:
q_water = m_water * c_water * ΔT_water
Given:
m_water = 50 g
c_water = 4,186 J/(kg · K)
ΔT_water = 87.1°C - 88.6°C = -1.5°C (Note: We convert to Kelvin since we need the temperature in Kelvin scale)
Converting grams to kilograms:
m_water = 50 g ÷ 1000 = 0.05 kg
Converting Celsius to Kelvin:
ΔT_water = -1.5°C + 273.15 = 271.65 K
Plugging in the values:
q_water = 0.05 kg * 4,186 J/(kg · K) * 271.65 K
q_water = 5688.97 J
The heat gained by the water is 5688.97 J (rounded to the nearest hundredth).
Now, we know that the heat gained by the water is equal to the heat lost by the metal since it's an isolated system:
q_water = q_metal
Now, let's calculate the heat lost by the metal:
q_metal = m_metal * c_metal * ΔT_metal
Given:
m_metal = 12.9 g
ΔT_metal = 87.1°C - 26.5°C = 60.6°C (Again, we convert to Kelvin)
Converting grams to kilograms:
m_metal = 12.9 g ÷ 1000 = 0.0129 kg
Converting Celsius to Kelvin:
ΔT_metal = 60.6°C + 273.15 = 333.75 K
Plugging in the values:
q_metal = 0.0129 kg * c_metal * 333.75 K
Since we know that q_water = q_metal, we can set them equal to each other and solve for c_metal:
q_water = q_metal
5688.97 J = 0.0129 kg * c_metal * 333.75 K
Simplifying the equation:
c_metal = 5688.97 J / (0.0129 kg * 333.75 K)