Question

Introduction: The specific heat capacity of a substance is the amount of energy needed to change the temperature of that substance by 1 °C. Specific heat capacity can be calculated using the following equation:
q = mc deltaT
In the equation q represents the amount of heat energy gained or lost in joules), m is the mass of the substance (in grams), c is the specific heat capacity of the substance (in J/g °C), and AT is the temperature change of the substance in °C).
Goal: Calculate the specific heat capacities of copper, granite, lead, and ice.
Solve: When you mix two substances, the heat gained by one substance is equal to the heat lost by the other substance. Suppose you place 125 g of aluminum in a calorimeter with 1,000 g of water. The water changes temperature by 2 °C and the aluminum changes temperature by -74.95 °C.
A. Water has a known specific heat capacity of 4.184 J/g °C. Use the specific heat equation to find out how much heat energy the water gained (q).
B. Assume that the heat energy gained by the water is equal to the heat energy lost by the aluminum. Use the specific heat equation to solve for the specific heat of aluminum. Aluminum's accepted specific heat value is 0.900 J/g °C. Use this value to check your work.

Answer 1

Answer:

A) 8,368 J

B) ) 0.893 J/gºC

Explanation:

A)

• The heat gained by the water can be obtained solving the following equation:

       $q_{g} = c_{w} * m * \Delta T (1)$

• where cw = specific heat of water = 4.184 J/gºC
• m= mass of water = 1,000 g
• ΔT = 2ºC
• Replacing these values in (1) we get:

       $q_{g} = c_{w} * m * \Delta T = 4.184 J/gºC*1,000 g* 2ºC = 8,368 J (2)$

B)

• Assuming that the heat energy gained by the water is equal to the one lost by the aluminum, we can use the same equation, taking into account that the energy is lost by the aluminum, so the sign is negative:  -8,368 J.
• Replacing by the mass of aluminum (125 g), and the change in temperature (-74.95ºC), in (1), we can solve for the specific heat of aluminum, as follows:

       $q_{l} = c_{Al} * m_{Al} * \Delta T (3)$

⇒    $-8,368 J = c_{Al}* 125 g * (-74.95ºC) (4)$

       $c_{Al} = \frac{-8,368J}{125g*(-74.95ºC} = 0.893 J/gºC (5)$

• which is pretty close to the Aluminum's accepted specific heat value of 0.900 J/gºC.