Question

A 235-g sample of a substance is heated to 350 ∘C and then plunged into a 105-g aluminum calorimeter cup containing 195 g of water and a 17-g glass thermometer at 11.5 ∘C. The final temperature is 35.0∘C. The value of specific heat for aluminium is 900 J/kg⋅C∘ , for glass is 840 J/kg⋅C∘ , and for water is 4186 J/kg⋅C∘What is the specific heat of the substance?

Answer 1

Alright, in order to solve for the specific heat of the substance, we will first need to calculate the amount of heat gained by the calorimeter cup, the thermometer, and the water, from the heat lost by the substance. Then, we will use that total gained heat to calculate the specific heat of the substance. Let's start.

1. We have the calorimeter cup which is made of aluminum with a mass of 105 grams (or 0.105 kg, because we need to use SI units), an initial temperature of 11.5 degrees Celsius, a final temperature of 35.0 degrees Celsius, and a specific heat capacity of 900 J/(kg*C). We will use the formula for heat,
q=mcΔT
where q is heat, m is mass, c is specific heat, and ΔT is the change in temperature (final temperature - initial temperature).
So for the calorimeter cup, q = 0.105 * 900 * (35.0 - 11.5) = 2220.75 J.

2. We apply the same process to the thermometer (17 g, or 0.017 kg) and the water (195 g, or 0.195 kg), with both having the same initial and final temperatures as the cup, 11.5 degrees Celsius and 35.0 degrees Celsius respectively, and specific heat capacities of 840 J/(kg*C) for glass (which we assume the thermometer is made of) and 4186 J/(kg*C) for water.
For the thermometer, q = 0.017 * 840 * (35.0 - 11.5) = 335.58 J.
For the water, q = 0.195 * 4186 * (35.0 - 11.5) = 19182.345 J.

3. The sum of total heat gained by all components is given by adding together the heats calculated in the above steps. Hence, the total heat gained is:
q_total = 2220.75 J + 335.58 J + 19182.345 J = 21738.675 J.

4. Now that we have calculated the total amount of heat gained (which is equivalent to the total heat lost by the sample), we can calculate the specific heat of the sample. The sample has a mass of 235 g (or 0.235 kg) and a temperature change from 350 degrees Celsius to 35.0 degrees Celsius.

5. To calculate the specific heat of the sample, we need to use the heat equation again, but this time we rearrange it to solve for c, specific heat:
c = q / ( m * ΔT)
By substituting the values, we'll get the specific heat of the substance:
c = 21738.675 J / ( 0.235 kg * (350 C - 35.0 C) = 293.67 J/(kg*C).

So, the specific heat of the substance that was initially heated is 293.67 J/(kg*C).