Answer:
40.02 calories
Explanation:
V = 10 mL = 10g
we know t went <em>up</em> by 4°C, this is our ∆t as it is a change.
Formula that ties it together: Q = mc∆t
where,
Q = energy absorbed by water
m = mass of water
c = specific heat of water (constant)
∆t = temperature change
Q = (10 g) x (4.186 J/g•°C) x (4°C)
Q = 167.44 J
Joules to Calories:
167.44 J x 1 cal/4.184 J = 40.02 calories
(makes sense as in image it is close to the value).
Answer:
Explanation:
In this problem,
we will determine the enthalpy at the end of heat input and we get the condition is the super heated state.
so the entropy change will be due to latent heat at 150 kPa and also due to temperature change at the super heated state.
All the temperature, enthalpy is evaluated with the help of the steam table.
mass is calculated with the help of a specific volume at initial condition.
<u>see image below</u>
