A 300g lead is heated to 95celsius and is then dropped into 100g of water at 10celsius. If the final temperature is 15celsius. What is the specific heat of the lead?
To find the specific heat of lead, we can use the principle of heat transfer. The heat gained by the lead is equal to the heat lost by the water. The formula for heat transfer is:
Q = m * c * ΔT
Where:
Q = heat transferred (in Joules)
m = mass (in kilograms)
c = specific heat capacity (in Joules per kilogram per degree Celsius)
ΔT = change in temperature (in degrees Celsius)
First, let's calculate the heat gained by the lead. Given:
Mass of lead, m_lead = 300g = 0.3kg
Initial temperature of lead, T_lead_initial = 95°C
Final temperature of lead and water, T_final = 15°C
Change in temperature for lead, ΔT_lead = T_final - T_lead_initial = 15°C - 95°C = -80°C
Now, let's calculate the heat lost by the water. Given:
Mass of water, m_water = 100g = 0.1kg
Initial temperature of water, T_water_initial = 10°C
Change in temperature for water, ΔT_water = T_final - T_water_initial = 15°C - 10°C = 5°C
Since the heat gained by the lead is equal to the heat lost by the water, we can set up the equation:
m_lead * c_lead * ΔT_lead = m_water * c_water * ΔT_water
Substituting the given values:
0.3kg * c_lead * (-80°C) = 0.1kg * c_water * 5°C
We know that the specific heat capacity of water is approximately 4186 J/(kg·°C). Now we need to solve for the specific heat capacity of lead, c_lead.
0.3kg * c_lead * (-80°C) = 0.1kg * 4186 J/(kg·°C) * 5°C
Rearranging the equation to solve for c_lead:
c_lead = (0.1kg * 4186 J/(kg·°C) * 5°C) / (0.3kg * -80°C)
c_lead ≈ -139.5 J/(kg·°C)
Therefore, the specific heat of lead is approximately -139.5 J/(kg·°C).
Note: The negative sign indicates that lead loses heat while heating, rather than gaining it, which is consistent with our calculation result.