Question

When you mix two substances, the heat gained by one substance is equalto the heat lost by the other substance. Suppose you place 125 g of aluminum ina calorimeter with 1,000 g of water. The water changes temperature by 2 °C andthe aluminum changes temperature by â74.95 °C.

Required:
a. Water has a known specific heat capacity of 4.184 J/g °C. Usethe specific heat equation to find out how much heat energy the watergained (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. (Hint: Because heat energy is lost, the value of q is negative.)

Answer 1

A: the heat gained is 8368 J (8.368 KJ)

B: 0.893 J/g°C

Step-by-step explanation:

A:

q = ?

m = 1000 g

C = 4.184 J/g°C

△T = 2°C

q = mc△T

q = (1000g) (4.184 J/g °C) (2 °C)

q = 8368 J

B:

q = mc△T

---------------

△T △T

c = q / m△T

c = - 8368 J / (125 g) (-74.95 °C)

c = 0.893 J/g°C

Answer 2

A) 8,368 J

B) 0.893 J/gºC

Step-by-step 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) = (-8,368J)/(125g*(-74.95ºC) = 0.893 J/gºC (5)`