3. How many joules are required to raise the temperature of a 29.352g sample of water by 52.0°C? (the specific heat of water is 4.184 J/g °C)​

6,386.05 J are required to raise the temperature of a 29.352g sample of water by 52.0°C.

Explanation:

Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.

There is a direct proportional relationship between heat and temperature. The constant of proportionality depends on the substance that constitutes the body as on its mass, and is the product of the specific heat by the mass of the body. So, the equation that allows calculating heat exchanges is:

Q = c * m * ΔT

where Q is the heat exchanged by a body of mass m, made up of a specific heat substance c and where ΔT is the temperature variation.

In this case, you know:

Replacing:

Q= 4.184 *29.352 g* 52 C

Solving you  get:

Q= 6,386.05 J

6,386.05 J are required to raise the temperature of a 29.352g sample of water by 52.0°C.

hey there!

mass = 6.75 g

volume =   9 cm³

therefore:

d = mass / volume

d   = 6.75 / 9

d = 0.7 g/cm³

answer b