120 grams of metal at 88 degree celcius is poured into a 70 grams isolated cup contains 250 grams of water at 16 degree celcius. The final temperature of the mixture is 17.5 degree celcius. Determine the specific heat capacity of the metal?
To determine the specific heat capacity of the metal, we can use the principle of heat transfer, which states that the heat lost by the metal is equal to the heat gained by the water and cup.
Step 1: Calculate the heat lost by the metal.
The heat lost by the metal can be calculated using the formula:
Q = m * c * ΔT
where Q is the heat lost, m is the mass of the metal, c is the specific heat capacity of the metal, and ΔT is the change in temperature.
Given:
m_metal = 120 grams
ΔT_metal = 88°C - 17.5°C = 70.5°C
Step 2: Calculate the heat gained by the water and cup.
The heat gained by the water and cup can also be calculated using the formula:
Q = m * c * ΔT
where Q is the heat gained, m is the combined mass of the water and cup, c is the specific heat capacity of water (which is approximately 4.18 J/g°C), and ΔT is the change in temperature.
Given:
m_water = 250 grams
m_cup = 70 grams
c_water = 4.18 J/g°C
ΔT_water = 17.5°C - 16°C = 1.5°C
Total mass of the water and cup:
m_total = m_water + m_cup = 250 grams + 70 grams = 320 grams
Step 3: Equate the heat lost by the metal to the heat gained by the water and cup.
Q_metal = Q_water + Q_cup
m_metal * c_metal * ΔT_metal = m_total * c_water * ΔT_water
Substituting the known values:
120g * c_metal * 70.5°C = 320g * 4.18 J/g°C * 1.5°C
Step 4: Solve for the specific heat capacity of the metal (c_metal).
Dividing both sides of the equation by (120g * 70.5°C), and rearranging the equation, we get:
c_metal = (320g * 4.18 J/g°C * 1.5°C) / (120g * 70.5°C)
Simplifying:
c_metal ≈ 0.418 J/g°C
Therefore, the specific heat capacity of the metal is approximately 0.418 J/g°C.