Question

1)How much internal energy is generated when a 18.5-g lead bullet, traveling at 6.30 x 10² m/s, comes to a stop as it strikes a metal plate?

2)What is the heat capacity of 29.1 kg of flint glass?

3)If 122.4 kJ of heat are supplied to 550 g of water at 20.7°C, what is the final temperature of the water? Give answer in °C.

4)An 76.4-kg man eats a banana of energy content 112 kcal. If all of the energy from the banana is converted into kinetic energy of the man, how fast is he moving, assuming he starts from rest?

5) a- What is the heat capacity of 1.83 m³ of mercury (Density = 13600 kg/m³)?

b- What is the heat capacity of 1.83 m³ of iron (Density = 7860 kg/m³)?

6)It takes 975 J to raise the temperature of 126 g of zinc from 0 to 20°C. What is the specific heat of zinc? Give answer in kJ/(kg·K).

7)A heating coil inside an electric kettle delivers 2.56 kW of electric power to the water in the kettle. How long will it take to raise the temperature of 0.577 kg of water from 23.1°C to 100°C?

Answer 1

1.solution

KE= - PE = internal energy of an object

KE = 0.5mv^2
KE = (0.5)(18.5*10^-3kg)(6.30x10²)^2
KE = 3671.32 J
Internal Energy = 3671.32 J

3.solution

specific heat of water = 4.184 J/gC

heat = mass* specific heat of water * change in temp
Q = M*s*T

find T.

Tf - Ti = Q / M*s

Tf - Ti =  122.4*10³ J / 550g x4.184 J/gC

Tf - Ti = 53.18 C

Tf = 53.18 + 20.7 = 73.88 ^oC

4.solution

Conserve energy:

KEi = KEf

Here, before is the banana and the man at rest, after is the man moving after eating the banana.

112 kcal = 112 x 4180 J (there are three different definitions of calorie - but they're all fairly similar numerically)

= 4.68*10⁵ J

if this is the kinetic energy of a 76.4 kg man

1/2 m*v^2 = 4.68*10⁵ J

v^2 = (2* 4.68*10⁵ ) / 76.4 = 110.70 m/s

v = 110.70 m/s

7.solution

specific heat of water is 4.186 kJ/kgC

E = 4.186 kJ/kgC x 0.577 kg x 76.9 C = 185.73 kJ
2.56 kW = 2.56 kJ/s
185.73 kJ / 2.56 kJ/s = 72.55 seconds