An 11.98-gram sample of zinc metal is placed in a hot water bath and warmed to 78.4°C. It is then removed and placed into a Styrofoam cup containing 50.0 mL of room temperature water (T=27.0°C; density = 1.00 g/mL). The water warms to a temperature of 28.1°C. Determine the specific heat capacity of the zinc
To determine the specific heat capacity of zinc, we need to use the equation:
q = m * c * ΔT
where:
q = heat gained or lost by the substance (in this case, water)
m = mass of the substance (water)
c = specific heat capacity of the substance (water)
ΔT = change in temperature
First, let's find the heat gained by the water:
q_water = m_water * c_water * ΔT_water
To find the mass of the water, we can use its density:
m_water = volume * density_water = 50.0 mL * 1.00 g/mL = 50.0 g
Now, let's calculate the change in temperature of the water:
ΔT_water = T_final - T_initial = 28.1°C - 27.0°C = 1.1°C
Plugging in these values:
q_water = 50.0 g * c_water * 1.1°C
Next, let's calculate the heat lost by the zinc:
q_zinc = m_zinc * c_zinc * ΔT_zinc
The change in temperature for the zinc can be calculated by:
ΔT_zinc = T_final - T_initial = 28.1°C - 78.4°C = -50.3°C
Now, let's substitute the known values:
q_zinc = 11.98 g * c_zinc * -50.3°C
Since the heat gained by the water is equal to the heat lost by the zinc:
q_water = q_zinc
We can now set up an equation and solve for c_zinc:
50.0 g * c_water * 1.1°C = 11.98 g * c_zinc * -50.3°C
Simplifying the equation:
c_zinc = (50.0 g * c_water * 1.1°C) / (11.98 g * -50.3°C)
c_zinc = (55.0 g * c_water) / (-551.23 g°C)
Finally, we can calculate the specific heat capacity of zinc:
c_zinc = (-0.1 * c_water) g°C
Please note that the specific heat capacity of water is typically around 4.18 J/g°C. You can substitute this value into the equation to find c_zinc.
To determine the specific heat capacity of zinc, we can use the formula:
q = m × c × ΔT
Where:
q = heat transfer
m = mass of the substance
c = specific heat capacity
ΔT = change in temperature
First, let's calculate the heat transfer (q) for water using the formula:
q_water = m_water × c_water × ΔT_water
Given:
m_water = 50.0 mL = 50.0 g (since the density of water is 1.00 g/mL)
c_water = specific heat capacity of water = 4.18 J/g°C (approximate value)
ΔT_water = (28.1°C - 27.0°C) = 1.1°C
q_water = (50.0 g) × (4.18 J/g°C) × (1.1°C)
q_water = 230.9 J
Next, let's calculate the heat transfer (q) for zinc using the formula:
q_zinc = m_zinc × c_zinc × ΔT_zinc
Given:
m_zinc = 11.98 g (mass of the zinc sample)
ΔT_zinc = (28.1°C - 78.4°C) = -50.3°C
q_zinc = (11.98 g) × (c_zinc) × (-50.3°C)
Since the zinc is losing heat to the water, the heat transfer should be negative.
Now, let's calculate the specific heat capacity of zinc (c_zinc) by rearranging the formula to solve for c_zinc:
c_zinc = q_zinc / (m_zinc × ΔT_zinc)
c_zinc = (-q_zinc) / (m_zinc × ΔT_zinc)
Substituting the known values:
c_zinc = (-230.9 J) / (11.98 g × -50.3°C)
c_zinc ≈ 0.386 J/g°C
Therefore, the specific heat capacity of zinc is approximately 0.386 J/g°C.
Sum of the heats gained is zero in the cup.
Heatgained by water+heatgained by zinc=0
50cw*(28.1-27)+11.98Czinc*(28.1-78.4)=0
solve for Czinc