118 g of a metal at 82.0°C are added to 50.0 g of water at 31.1°C. When the system reaches constant temperature, the temperature is 40.0°C. What is the specific heat of the metal? The specific heat of water is 4.184 J/g·°C.
Refer to your earlier post. Same procedure.
To find the specific heat of the metal, we can use the equation:
Q = mcΔT
Where:
Q is the heat absorbed or released by the substance
m is the mass of the substance
c is the specific heat of the substance
ΔT is the change in temperature
In this case, we want to find the specific heat of the metal. We know the mass of the metal (118 g), the initial temperature of the metal (82.0°C), the final temperature of the system (40.0°C), and the specific heat of water (4.184 J/g·°C).
First, let's calculate the heat absorbed by the water. Since the final temperature is 40.0°C, the change in temperature for the water is:
ΔT_water = 40.0°C - 31.1°C = 8.9°C
Using the equation for Q, we can calculate the heat absorbed by the water:
Q_water = m_water * c_water * ΔT_water
= 50.0 g * 4.184 J/g·°C * 8.9°C
= 1861.6 J
Now, let's calculate the heat released by the metal. Since the final temperature is the same for both the water and the metal, the change in temperature for the metal is:
ΔT_metal = 40.0°C - 82.0°C = -42.0°C
Using the equation for Q, we can calculate the heat released by the metal:
Q_metal = m_metal * c_metal * ΔT_metal
We want to find the specific heat of the metal, so we can rearrange the equation:
c_metal = Q_metal / (m_metal * ΔT_metal)
From the problem, we know that the heat released by the metal must be equal to the heat absorbed by the water, so Q_metal = Q_water:
c_metal = Q_water / (m_metal * ΔT_metal)
= 1861.6 J / (118 g * -42.0°C)
= -0.383 J/g·°C
The specific heat of the metal is approximately -0.383 J/g·°C. Note that the negative sign signifies that the metal released heat, whereas the positive sign would indicate absorption of heat.