A 15.0 g of sample of silver is heated to 100.0 0C and then placed in a constant-pressure calorimeter containing 25.0 g of water at 23.0 0C. The final temperature of the system was measured as 25.0 0C. What is the specific heat of silver?
The sum of the heats added is zero.
Heat gained by silver+heat gained by water=0
masssilver*Csilver*(25-100)C+masswater*cwater*(25-23)C=0
To find the specific heat of silver, we can use the formula:
q = m · c · ΔT
Where:
q = heat exchanged between the silver and water
m = mass of the substance (silver or water)
c = specific heat capacity of the substance
ΔT = change in temperature of the substance
First, let's find the heat exchanged between the silver and water using the formula:
q = m · c · ΔT
For the water:
m = 25.0 g (mass of water)
c = 4.18 J/g·°C (specific heat capacity of water)
ΔT = 25.0 °C - 23.0 °C = 2.0 °C
Now, we can calculate q for the water:
q_water = m_water · c_water · ΔT_water
q_water = 25.0 g · 4.18 J/g·°C · 2.0 °C
q_water = 209 J
Since q_water is equivalent to the heat gained by the water, the heat lost by the silver will be equal in magnitude but opposite in sign (due to heat conservation).
Now, let's find the heat lost by the silver:
q_silver = -q_water
q_silver = -209 J
Next, we can find the change in temperature (ΔT_silver) of the silver using the formula:
q_silver = m_silver · c_silver · ΔT_silver
Since we are given the final temperature as 25.0 °C and the initial temperature as 100.0 °C, the change in temperature can be calculated as:
ΔT_silver = 25.0 °C - 100.0 °C
ΔT_silver = -75.0 °C
Substituting the values into the equation:
q_silver = m_silver · c_silver · ΔT_silver
-209 J = 15.0 g · c_silver · -75.0 °C
Now, we can solve for c_silver (specific heat capacity of silver):
c_silver = -209 J / (15.0 g · -75.0 °C)
c_silver ≈ 0.186 J/g·°C
Therefore, the specific heat of silver is approximately 0.186 J/g·°C.