Question

The mass of the water to 1.0 kg (1000 g). The cylinder has a mass of 5.0 kg and a height of 500 m. The temperature changed from 25 degrees Celsius to 30.86 degrees Celsius in the experiment. Specific heat can be calculated using the following equation: q=mc(delta t)

What is the gravitational potential energy of the cylinder?
If no energy is lost, how much heat energy is added to the water?
What is the mass of the water?
What is the temperature change of the water?
What is the specific heat of the water?

Answer 1

1) 24500 J (29400 J if we include also the water)

The gravitational potential energy of an object is given by:

$U=mgh$

where m is the mass, g is the gravitational acceleration and h is the heigth above the ground.

Therefore, the gravitational potential energy of the cylinder is given by:

$U=mgh=(5.0 kg)(9.8 m/s^2)(500 m)=24500 J$

If we count also the water inside the cylinder, the total mass is 5.0 kg + 1.0 kg = 6.0 kg, and the total gravitational potential energy would be

$U=mgh=(6.0 kg)(9.8 m/s^2)(500 m)=29400 J$

2) 24530 J

The heat energy added to the water is given by:

$Q=mC\Delta T$

where

m = 1.0 kg is the mass of the water

C = 4186 J/kg C is the specific heat of the water

$\Delta T=30.86 C - 25 C=5.86 C$ is the increase in temperature

Substituing, we find

$Q=(1.0 kg)(4186 J/kgC)(5.86 C)=24530 J$

3) 1.0 kg

The mass of the water is given, and it is m = 1.0 kg.

4) $5.86^{\circ}C$

The temperature change of the water is the difference between its final temperature and its initial temperature:

$\Delta T=30.86 C -25 C=5.86^{\circ}C$

5)  $4186 J/kg ^{\circ}C$

The specific heat of the water is $4186 J/kg C$