1. Determine the specific heat of a certain metal if a 450-g sample of it loses 34,500 joules of heat as its temperature drops by 97degC.

2. The temperature of a 55-g sample of a certain metal drops by 113degC as it losses 3,500 J of heat. What is the specific heat of the metal?

3. A golfer wishes to hit his drives further by increasing the kinetic energy of the golf club when it strikes the ball. Which would have the greater effect on the energy transferred to the ball by the driver – doubling the mass of the club or doubling the speed of the club head? Explain.

4. A stone is dropped from a certain height and penetrates into the mud. All else being equal, if it is dropped from twice the height, how much farther should it penetrate?

1. To determine the specific heat of a certain metal, you can use the formula:

q = m * c * ΔT

where:
q = heat lost or gained by the metal (in joules)
m = mass of the metal (in grams)
c = specific heat capacity of the metal (in J/g°C)
ΔT = change in temperature of the metal (in °C)

In this case, you are given the mass of the metal (450 g), the heat lost (34,500 J), and the change in temperature (97°C). Substituting these values into the formula:

34,500 J = 450 g * c * 97°C

Divide both sides of the equation by (450 g * 97°C) to solve for c:

c = 34,500 J / (450 g * 97°C)
c ≈ 0.773 J/g°C

Therefore, the specific heat of the metal is approximately 0.773 J/g°C.

2. Similar to the previous question, you can use the same formula:

q = m * c * ΔT

Given the mass of the metal (55 g), the heat lost (3,500 J), and the change in temperature (113°C), substitute these values into the formula:

3,500 J = 55 g * c * 113°C

Divide both sides of the equation by (55 g * 113°C) to solve for c:

c = 3,500 J / (55 g * 113°C)
c ≈ 0.575 J/g°C

Therefore, the specific heat of the metal is approximately 0.575 J/g°C.

3. When it comes to increasing the energy transferred to the golf ball by the driver, doubling the speed of the club head would have a greater effect than doubling the mass of the club.

The kinetic energy (KE) of an object can be calculated using the formula:

KE = 0.5 * m * v^2

where:
m = mass of the club head
v = velocity of the club head

Doubling the mass of the club (m) would directly double the kinetic energy (KE), given that the velocity remains the same. However, doubling the speed of the club head (v) would significantly increase the kinetic energy. Since velocity has a squared relationship with kinetic energy, doubling the speed of the club head would result in four times the kinetic energy transfer to the ball.

Therefore, doubling the speed of the club head would have a greater effect on the energy transferred to the ball by the driver than doubling the mass of the club.

4. When a stone is dropped, its penetration into the mud depends on its initial velocity, which is determined by its height. Assuming all else is equal (no air resistance, same stone, and constant speed on impact), dropping the stone from twice the height would result in it having twice the initial velocity when it hits the mud.

The distance a stone will penetrate into the mud (d) can be approximated by considering the energy of the stone and the resistance provided by the mud. Since kinetic energy (KE) is directly related to the square of velocity (v^2), when the initial velocity is doubled, the kinetic energy increases by a factor of 4.

Therefore, if the stone is dropped from twice the height, it should penetrate into the mud approximately four times farther compared to when it was dropped from the original height.