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?
A 500g copper at 100celsius is placed in a 200g water at 25celsius which is contained in an aluminum calorimeter that has a mass of 400g. If the final temperature of the whole system is 30celsius, find the specific heat of the lead.
The sum of heats GAINED is zero.
heat Gained=mass*specificeheat*(Tf-Ti)
Heatgainedby water + heatgained by lead=0
Mwater*specifhtWater*(Tf-ti)+Masslead*specihtLead*(Tf-Ti)=0
solve.
To find the specific heat of the lead, we can use the principle of heat transfer:
The heat gained by the water = heat lost by the lead.
The amount of heat gained by the water can be calculated using the formula:
Q = mcΔT
where Q is the heat gained, m is the mass of the water, c is the specific heat of water, and ΔT is the temperature change of the water.
The heat lost by the lead can also be calculated using the same formula:
Q = mcΔT
where Q is the heat lost, m is the mass of the lead, c is the specific heat of the lead, and ΔT is the temperature change of the lead.
In this case, the final temperature of the mixture is 15 degrees Celsius. We can calculate the temperature change of the water and the lead from their initial temperatures to the final temperature:
Temperature change of water = (final temperature - initial temperature of water)
= (15°C - 10°C)
= 5°C
Temperature change of lead = (final temperature - initial temperature of lead)
= (15°C - 95°C)
= -80°C
Next, we can substitute the values into the formulas:
Heat gained by the water = (m_water)(c_water)(ΔT_water)
= (100g)(4.186 J/g°C)(5°C)
= 2093 J
Heat lost by the lead = (m_lead)(c_lead)(ΔT_lead)
= (300g)(c_lead)(-80°C)
= -24000c_lead
Since the total heat gained must equal the total heat lost, we can equate the two values:
2093 J = -24000c_lead
To find c_lead, we can rearrange the equation:
c_lead = 2093 J / -24000
Calculating this gives us:
c_lead ≈ -0.0872 J/g°C
The specific heat of lead is approximately -0.0872 J/g°C.