Answer Specific Heat

Answer: Specific Heat is: -the amount of energy required to raise the temperature of 1g of a substance 1°C -specific heat is symbolised as Cp or C -has units of J/g °C -J stands for Joules, which is a unit of energy Most often used in equation: q = mΔTCp q = energy in J m = mass in g ΔT = change in temperature Cp = specific heat

Example How much energy is required to raise the temperature of 345.34g of Aluminium from 35.0°C 35 .0°C to 250.00°C ? The specific heat of Aluminium is .90 J/Cp q = mΔTCp q = 345.34g(250.00°C - 35.0°C)0.09J/g°C from there its just simple Algebra. q = 66,823.29J

However, not all problems will be as simple as the one above. Some may require up to 3+ equations that can include Enthalpy or or ΔH which leads to using a Change Ch ange of State Graph.

Read more: http://wiki.answers.com/Q/How_do_you_solve_specific_heat_problems#ixzz1LwZfMBBr 

Learn It: Sample Problems - Specific Heat Capacities Exam tip: When solving problems involving specific heat capacities, you only need to know two equations!

1.

Heat = mass × specific heat capacity × temperature change Q = mc Dq

and

2.

Heat lost = Heat gained mc Dq = mc Dq

When reading the question, concentrate on what is losing heat and what is gaining heat - then use the above two equations to solve the problem.

Worked Examples Question 1

Given that the specific heat capacity of water is 11 times that of copper, calculate the mass of copper at a temperature of 100 °C required to raise the temperature of 200 g of water from 20.0 °C to 24.0 °C, assuming no energy is lost to the surroundings.

Solution

Heat lost by copper = heat gained by water mcuccuDqcu = mwcwDqw mcuccu(100 - 24) = 0.200 × 11ccu(24 – 20) 76mcu = 8.8 mcu = 0.116 kg

Question 2 Three litres of water at 100 °C are added to 15 litres of water at 40 °C. Calculate the temperature of the mixture. Take the mass of 1 litre of water to be 1 kg and the specific heat capacity of water to be 4.2 × 103 J kg

-1

K

-1

Solution

Let the temperature of the mixture q Heat lost = heat gained m1c1Dq1 = m2 c2 Dq2 3 × 4.2 × 103 × (100 – q) = 15 × 4.2 × 103 × (q -40) Solving for q gives q = 50°C

You try it!

Question 3

1 kg of water at a temperature of 45 °C is mixed with 1.5 kg of alcohol at 20 °C. Find the final temperature of the mixture.

Take the specific heat capacity of water to be 4200 J kg be 2400 J kg

–1

K

–1

–1

K

–1

and the specific heat capacity of alcohol to

. Assume no other exchange of heat occurs.

Solution

Let the final temperature of the mixture be q. Heat lost by water = heat gained by alcohol mwcwDqw = malcalDqal 1 × 4200 × (45 – q) = 1.5 × 2400 × (q – 20) (mcDq)w = (mcDq)al 1 × 4200 × (45 – q) = 1.5 × 2400 × (q – 20)

Solving for q gives q = 33 °C.

You try it!

Application of specific heat capacity: storage heaters Some electrical heaters contain oil which has a high boiling point and a relatively high specific heat capacity. Once the oil has been heated it retains the thermal energy for quite a while, cooling down only slowly.

In storage heaters, a core of high-density thermal blocks (like very hard building blocks), is heated by electric elements during the night. These blocks are preferred to water, despite having a slightly lower specific heat capacity, because they are more dense (and thus more compact) and do not leak. They are heated during the night using low-cost (off-peak) electricity and release the stored thermal energy during the day.

Storage heaters maintain a steady room air temperature over long periods, warming the entire building fabric and reducing the risk of condensation.