A 230-gram sample of metal at 82 degrees centigrade is added to 632 grams of water at 14.4 degrees centigrade in an ideal calorimeter. The final temperature of the system (water and metal) is 22.8 degrees centigrade. What is the heat acapacity of the metal. The specific heat of water is 4.18 J/(g • °C)
The sum of the heats added is zero (one will be negative added)
heataddedmetal+heatadded water=0
massmetal*cm*(Tf-82)_+ masswater*cw*(Tf-14.4)=0
solve for cm
To find the heat capacity of the metal, we can use the principle of heat transfer, which states that the heat gained by the water is equal to the heat lost by the metal.
The equation for heat transfer is:
Q = m * c * ΔT
Where:
Q is the heat
m is the mass
c is the specific heat
ΔT is the change in temperature
For the water, the heat gained (Q_water) can be calculated as:
Q_water = m_water * c_water * ΔT_water
For the metal, the heat lost (Q_metal) can be calculated as:
Q_metal = m_metal * c_metal * ΔT_metal
Since the two quantities of heat are equal, we can set them equal to each other and solve for the specific heat of the metal (c_metal):
m_water * c_water * ΔT_water = m_metal * c_metal * ΔT_metal
Plugging in the given values:
(632 g) * (4.18 J/(g • °C)) * (22.8°C - 14.4°C) = (230 g) * c_metal * (82°C - 22.8°C)
Simplifying the equation:
(26356.8 J) = (230 g) * c_metal * (59.2°C)
Dividing both sides of the equation by (230 g * 59.2°C), we get:
c_metal = (26356.8 J) / ((230 g) * (59.2°C))
Evaluating the expression:
c_metal ≈ 1.58 J/(g • °C)
Therefore, the heat capacity of the metal is approximately 1.58 J/(g • °C).