Answer:
heat, energy that is transferred from one body to another as the result of a difference in temperature. If two bodies at different temperatures are brought together, energy is transferred—i.e., heat flows—from the hotter body to the colder. example: stove
Explanation:
hope this helps
Q1. The answer is 8.788 m/s
V2 = V1 + at
V1 - the initial velocity
V2 - the final velocity
a - the acceleration
t - the time
We have:
V1 = 4.7 m/s
a = 0.73 m/s²
t = 5.6 s
V2 = ?
V2 = 4.7 + 0.73 * 5.6
V2 = 4.7 + 4.088
V2 = 8.788 m/s
Q2. The answer is 9.22 s
V2 = V1 + at
V1 - the initial velocity
V2 - the final velocity
a - the acceleration
t - the time
We have:
V2 = 0 (because it reaches a complete stop)
V1 = 4.7 m/s
a = -0.51 m/s²
t = ?
0 = 4.7 + (-0.51)*t
0 = 4.7 - 0.51t
0.51t = 4.7
t = 4.7 / 0.51
t = 9.22 s
Answer:
Hope it helps you :)
Explanation in the pic above.
Answer:
The amount of kilograms of ice at -20.0°C that must be dropped into the water to make the final temperature of the system 40.0°C = 0.0674 kg
Explanation:
Heat gained by ice in taking the total temperature to 40°C = Heat lost by the water
Total Heat gained by ice = Heat used by ice to move from -20°C to 0°C + Heat used to melt at 0°C + Heat used to reach 40°C from 0°C
To do this, we require the specific heat capacity of ice, latent heat of ice and the specific heat capacity of water. All will be obtained from literature.
Specific heat capacity of ice = Cᵢ = 2108 J/kg.°C
Latent heat of ice = L = 334000 J/kg
Specific heat capacity of water = C = 4186 J/kg.°C
Heat gained by ice in taking the total temperature to 40°C = mCᵢ ΔT + mL + mC ΔT = m(2108)(0 - (-20)) + m(334000) + m(4186)(40 - 0) = 42160m + 334000m + 167440m = 543600 m
Heat lost by water = mC ΔT = 0.25 (4186)(75 - 40) = 36627.5 J
543600 m = 36627.5
m = 0.0674 kg = 67.4 g of ice.
