Answer:
2.55 × 10³ J =2.55 kJ
Explanation:
Specific heat capacity of ice = 37.8 J / mol °C
Specific heat capacity of water = 76.0 J/ mol °C
Ice at -12 °C is converted to ice at 0 °C by absorbing heat Q₁
Ice at 0°C melts to water at 0 °C. Let Heat absorbed during this phase change be Q₂ .
Let heat absorbed to raise the temperature of water from 0 C to 24°C be Q₃ .
Total heat = Q = Q₁ + Q₂ + Q₃
Q₁ = (37.8 j/mol C )(5.53 g /18.01532 g/ mol )( 0-(-12)) = 139.23749 j
Q₂ =(5.53 g/18.01532 g H₂O / mol ) (6.02 x10³ j) = 1847.905 j
Q₃ = (76 j/mol C) ( (5.53 g/18.01532 g H₂O / mol )(24-0) = 559.8968 j
Total Heat required = Q = 139.23749 j + 1847.905 j + 559.8968 j
= 2547.039 j = 2.55 × 10³ J =2.55 kJ
Answer:
Explanation:
Given:
- mass of water,
- initial temperature of water,
- initial temperature of pan,
- mass of pan,
- mass of water evapourated,
- specific heat of water,
- specific heat of aluminium pan,
- latent heat of vapourization,
<u>Using the equation of heat:</u>
<em>Here, initially certain mass of water is vapourised first and then the remaining mass of water comes in thermal equilibrium with the pan.</em>
Answer:
O D.
Explanation:
Physics has an aspect that deals with the study of energy
Answer:
B. 6
Explanation:
i think... im in 7th grade and haven't really leaned this but im like 60% sure but i migjt be wrong
Answer:
The weight limit of 300kg is the maximum amount the machine can handle so it can be dangerous to exceed the maximum load.
