A 215 g sample of a substance is heated to 330 degrees C and then plunged into a 105 g aluminum calorimeter cup containing 165 g of water and a 17 g glass thermometer at 12.5 degrees C. The final temp is 35.0 degrees C. What is the specific heat of the substance?

To find the specific heat of the substance, we can make use of the equation:

q = m * c * ΔT

Where:
q is the heat transferred,
m is the mass of the substance or the specific heat capacity of the substance,
c is the specific heat capacity of the substance,
ΔT is the change in temperature.

First, let's find the heat transferred from the substance to the water:

q1 = m1 * c1 * ΔT1

Where:
m1 is the mass of the substance,
c1 is the specific heat capacity of the substance,
ΔT1 is the change in temperature.

We are given:
m1 = 215 g
ΔT1 = 35.0 - 330 = -295.0 degrees C (Note: the change in temperature is negative since the substance is losing heat)

Now, let's find the heat transferred from the water:

q2 = m2 * c2 * ΔT2

Where:
m2 is the mass of the water,
c2 is the specific heat capacity of water (4.18 J/g·°C)
ΔT2 is the change in temperature.

We are given:
m2 = 165 g + 17 g + 105 g = 287 g
ΔT2 = 35.0 - 12.5 = 22.5 degrees C

The total heat transferred, q, can be calculated by adding q1 and q2:

q = q1 + q2

Now, substituting the values into the equation:

q = m1 * c1 * ΔT1 + m2 * c2 * ΔT2

We can solve for c1, the specific heat capacity of the substance:

c1 = q / (m1 * ΔT1)

Finally, let's substitute the given values into the equation:

c1 = q / (215 g * -295.0 degrees C)

To find the specific heat of the substance, we need to use the principle of conservation of energy and the formula for heat transfer:

Q(substance) + Q(aluminum cup) + Q(water) + Q(thermometer) = 0

First, let's calculate Q(substance), which is the heat gained or lost by the substance:

Q(substance) = mass(substance) * specific heat(substance) * change in temperature(substance)

Given:
mass(substance) = 215 g
change in temperature(substance) = final temperature - initial temperature
= 35.0 degrees C - 330 degrees C

Now let's calculate Q(aluminum cup), which is the heat gained or lost by the aluminum cup:

Q(aluminum cup) = mass(aluminum cup) * specific heat(aluminum) * change in temperature(aluminum cup)

Given:
mass(aluminum cup) = 105 g
specific heat(aluminum) = 0.897 J/g°C (specific heat of aluminum)
change in temperature(aluminum cup) = final temperature - initial temperature
= 35.0 degrees C - 12.5 degrees C

Next, let's calculate Q(water), which is the heat gained or lost by the water:

Q(water) = mass(water) * specific heat(water) * change in temperature(water)

Given:
mass(water) = 165 g
specific heat(water) = 4.184 J/g°C (specific heat of water)
change in temperature(water) = final temperature - initial temperature
= 35.0 degrees C - 12.5 degrees C

Finally, let's calculate Q(thermometer), which is the heat gained or lost by the thermometer:

Q(thermometer) = mass(thermometer) * specific heat(glass) * change in temperature(thermometer)

Given:
mass(thermometer) = 17 g
specific heat(glass) = 0.84 J/g°C (specific heat of glass)
change in temperature(thermometer) = final temperature - initial temperature
= 35.0 degrees C - 12.5 degrees C

Now, substitute the given values and calculate the heat transfer for each component.

Once we have calculated the heat transfer for each component, we can substitute the values into the conservation of energy equation and solve for the specific heat of the substance.

Please show some effort of your own other than changing a letter in your name each time you post.

(1) Look up the specific heats of glass, aluminum and water.

(2) Compute the amount of heat energy gained by those three materials as they heat from 12.5 to 35 C. Call that heat Q.
It equals the heat lost by the sample.

(3) Q = M(sample)*C(sample)*295C
Solve for the specific heat C of the sample.