3.90g sample of Al with an unknown initial temperature is dropped into 10 mg of water at 22.6 degrees Celsius. What will the change in temperature be of the Al if its final temperature is 28.6 degrees Celsius?

To find the change in temperature of the Al sample, we can use the equation:

q = mcΔT

Where:
q is the heat gained or lost by the substance (in this case, Al)
m is the mass of the substance (in grams)
c is the specific heat capacity of the substance (in J/g°C)
ΔT is the change in temperature (in °C)

First, we need to determine the heat gained or lost by the Al:

q = mcΔT

Since the Al sample is dropping in temperature, the heat gained by water will be equal to the heat lost by Al:

q(Al) = -q(water)

Next, we calculate the heat gained by water:

q(water) = mcΔT

Where:
m is the mass of the water (in grams)
c is the specific heat capacity of water (4.184 J/g°C)
ΔT is the change in temperature of the water (final temperature - initial temperature)

Given:
Mass of Al (m_Al) = 3.90 g
Specific heat capacity of Al (c_Al) = 0.897 J/g°C (approximate value)
Mass of water (m_water) = 10 mg = 0.01 g
Specific heat capacity of water (c_water) = 4.184 J/g°C
Initial temperature of water (T_initial_water) = 22.6 °C
Final temperature of water (T_final_water) = 28.6 °C

First, let's calculate the change in temperature of the water:

ΔT_water = T_final_water - T_initial_water
= 28.6 °C - 22.6 °C
= 6 °C

Next, let's calculate the heat gained by water:

q_water = m_water * c_water * ΔT_water
= 0.01 g * 4.184 J/g°C * 6 °C
= 0.25044 J

Since the heat gained by the water is equal to the heat lost by the Al, we can equate the two equations:

q_Al = -q_water

m_Al * c_Al * ΔT_Al = -q_water

ΔT_Al = -q_water / (m_Al * c_Al)

Now, we substitute the known values:

ΔT_Al = -0.25044 J / (3.90 g * 0.897 J/g°C)
= -0.0835 °C

Since the change in temperature is a decrease in this case, the final change in temperature of the Al sample will be -0.0835 °C.