
[Ended]Physics Tutoring Session #7 Measurement of Heat Energy on Mon. 29th Feb., 2016 @ 4:00  5:00pm
Physics
Chima Sunday (Tutor)
29022016 12:17:00 +0000The concept Measurement of Heat Energy is one of the fundamentals in Thermal Physics, often tested in JAMB and other exams. so lets take a quick look on this topics. Come along with your questions. and were you need more explanations.
Welcome and enjoy your time. This section has endedMeasurement of Heat Energy
Heat is a form of energy called thermal energy that flows due to temperature difference.
Specific Heat Capacity of Materials
recall that Q ∝ mδθ removing proportionality
∴ Q = cmδθ
where Q is the quantity of heat, m is the mass, δθ is the change in temperature and c is the specific heat capacity of the material.
Thus Specific heat capacity of a substance is the quantity of heat required to raise temperature of unit mass (1kg) of a substance through a degree rise in temperature(i.e 1^{o}C or 1k).
Heat Capacity H
The quantity of heat required to raise the temperature of the entire body through one degree rise in temperature(1k)
H = mc,
H = heat capacity, m = mass, c = specific heat capacity.
We can measure the specific heat capacity in two methods (a) method of mixture (2)Electrical method.
Latent Heat.
Latent heat of Fusion is the heat energy required to convert a substance from its solid form to its liquid form without any change in temperature.When a unit mass of that substance is to be considered, we talk of the specific latent heat of fusion (L)
thus L_{f} = Q/m. its unit is in j/kg.
Also, the specific latent heat of vapourisation of a substance is the quantity of heat required to change a unit mass of substance from liquid to vapour without any change in temperature.
the formula still holds L_{v}= Q/m
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Okon Daniel
Equal masses of two liquids X and Y at temperatures 30°c and 80°c respectively are mixed such that the temperature of the mixture is T. If the specific heat of X is twice that of Y,the value of T is___________?
0 29022016 12:53:00 +0000

Chima Sunday (Tutor)
Yah, once again you are specially welcome to today's tutorial, promised to make wort the time. Feel free to ask all your questions while i will try within my power to give you a satisfactory answer. I'm pleased to have you here. Welcome
0 29022016 15:03:00 +0000

Chima Sunday (Tutor)
@Okon Daniel, recall that if there is no heat loss to the environment, then heat gain by the one on lower temperature is equal to heat loss by the one on higher temerature.
i.e Heat Gain by X= Heat Loss By Y
so heat gain by X = Q_{X} = m_{X}c_{X}(T  30)
also Heat loss by Y = Q_{y} = m_{y}c_{y}(80  T)
from the question, m_{y} = m_{x}and c_{X} = 2c_{y}
so heat gain = heat loss
Q_{x} = Q_{y}
m_{y}2c_{y}(T  30) = m_{y}c_{y}(80  T)
by cancelling
2T  60 = 80  T
3T = 140
T = 46.7^{o<\small>}0 29022016 15:20:00 +0000

Chima Sunday (Tutor)
Hello are we still there can any one try this one
Jamb 2015 Question 14 A block of aluminum of mass m is heated electrically by 25W heater. if the temperature rises by 10oC in 5 minutes. Calculate the heat capacity of the aluminum.
Option A 750JK1
Option B 1250JK1
Option C 125JK1
Option D 5OJK1
updated solution work done by heater = heat gain by metal Solution Here, m is the mass of Aluminum Temperature T =10°C Time t = 5 mins. = 5 × 60 = 300s Heat absorbed = electrical energy m × c × T = P × t Where c is the specific heat capacity and mc = H(Heat capacity) H × 10 = 25 × 300 H = 25 * 300/10 = 25 * 30 = 750J/k Answer is option A0 29022016 15:24:00 +0000

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