The final temperature of a block of iron whose mass is 75.0 g that was initially at 22 °C after gaining heat during condensation of water of mass 0.95 g is <u>90.8 °C.</u>
<h2>Further Explanation:</h2><h3>Heat capacity </h3>
- Heat capacity refers to the amount of heat that is required to raise the temperature of an object or a substance by one degree Celsius or 1 Kelvin.
- Heat capacity is thus expressed as Joules/°C
<h3>Specific heat capacity </h3>
- This is the amount of heat that is required to raise the temperature of a unit mass of a substance by 1 degree Celsius or 1 Kelvin.
- It is expressed as Joule/kg/°C or J/g/°C. For example the specific heat of water is 4.184 j/g/°C
<h3>Quantity of Heat </h3>
- Quantity of Heat is calculated using the formula;
Q = mCΔT
Where Q is the Quantity of heat, m is the mass of the object, C is the specific heat capacity of the object and ΔT is the resulting change in temperature.
- ΔT is given by the difference between initial temperature and final temperature.
That is; ΔT = Tfinal - Tinitial
<h3>In this Question;</h3>
Heat released during condensation is equivalent to the heat absorbed by iron block.
<h3>Step 1; Moles of water </h3>
1 mole of water is 18.0 g/mol
Thus; number of moles = 0.95 g/18 g/mol
= 0.053 moles of water
Step 2: Heat released during condensation
Heat released during condensation = Number of moles x Latent heat of vaporization
Q = 0.053 moles x 44.0 kJ/mol
= 2.322 Kj
<h3>Step 3: Heat gained by Iron block and change in temperature</h3>
Q = mCΔT
m = 75 g
c = 0.450 J/g/°C
Therefore;
Q = 75 g x 0.450 J/g/°C x ΔT
2322 J = 75 g x 0.450 J/g/°C x ΔT
ΔT = 68.8°C
<h3>Step 4; Calculating final temperature </h3>
ΔT = Tfinal - Tinitial
Initial temperature = 22°C
Final temperature = initial temperature + Change in temperature
= 68.8 + 22
<u>= 90.8 °C</u>
<u></u>
Keywords: Change in temperature, Specific heat capacity, Quantity of heat
<h3>Learn more about;</h3>
Level: High school
Subject; Physics
Topic: Quantity of Heat