Specific heat capacity Nelson p STAWA p26

Specific heat capacity If you boil just enough water for one cup of tea in a kettle it takes a lot less time than it takes to boil a full kettle. What does this tell you? The amount of energy required to increase the temperature of an object is proportional to its mass.

Specific heat capacity To make nice hot soup you need to leave it on the stove for longer. More time = more energy. The amount of energy transferred to an object is proportional to the rise in temperature.

Specific heat capacity A metal spoon used to stir your soup will heat up a lot faster than a wooden spoon. Both spoons receive the same amount of energy. The amount of energy required to increase the temperature of an object is dependent on the material

When adding heat to an object the rise in temperature will depend on three things: (a) the amount of heat energy that you put in (b) the amount (mass) of the object that you are heating (c) the material from which the object is made. This is referred to as a materials heat capacity. Specific heat capacity

Specific heat capacity, c, is the energy required to raise 1 kg of a material by 1 K (or 1°C). units: J kg -1 K -1 or J kg -1 °C -1 (See STAWA page 26) Specific heat capacity Formula

To calculate the transfer of energy required for a particular temperate change we use: Specific heat capacity ΔQmcΔTΔQmcΔT heat energy transferred (J) mass (kg) specific heat capacity (J kg -1 K -1 ) change in temperature (K or °C)

How much energy does it take to raise the temperature of 1.5 litres of water by 20 ° C? m water = 1.5 L ( 1kg / 1L) = 1.5 kg c water = 4180 J kg -1 K -1  T = 20 ° C ΔQ = 1.5 kg x 4180 J kg -1 K -1 x 20 ° C = J Specific heat capacity – example 1 = 125 kJ

A Hot Jogger In 30 minutes, a 65 kg jogger can generate 8.0x10 5 J of heat. This heat is removed from the body by a variety of means, including the body’s own temperature-regulating mechanisms. If the heat were not removed, how much would the body temperature increase? Specific heat capacity – example 2 = 3.52°C

If there is no heat loss to the surroundings (in a closed system), the heat lost by the hotter object equals the heat gained by the cooler ones. Thermal equilibrium

Heat lost by copper = heat gained by water Thermal equilibrium

A 20g lead nut at a temperature of 80°C is placed in a polystyrene cup holding 250ml of water at a temperature of 22°C. What is the final temperature of the water? (Ignore any heat lost to the cup or surroundings.) Thermal equilibrium - example Soln: 22.1°C

Super heated pie filling! Some high temperature foods, you can eat comfortably when you take them out of the oven as they have a low specific heat capacity and therefore don’t hold much thermal energy but some foods like pie filling you can burn your mouth on as the high temperature food will hold a lot of energy. Hot water bottles A hot water bottle contains boiling water that cools gradually during the night releasing a large amount of thermal energy. Warming effects of the sea Countries surrounded by water are heated by the warm winds that have absorbed thermal energy from the ocean. The ocean cools gradually during the winter so maintains a constant source of heat energy. The water acts as a temperature moderator, absorbing energy from the air above in the summer and releasing it in the winter. Specific heat - examples

Why does the pie filling and not the pastry burn your mouth? They have both been sat in the oven for long enough to come to thermal equilibrium – same average E k 1. Thermal conductivity: pastry has lots of air, pie filling is more dense and contains water 2. The pie filling has a higher SHC than the crust: - both have risen to the same temperature but more energy is required by the filling to get there - the filling must give off a lot of heat for its temperature to decrease to that of your tongue

We can think of specific heat as “thermal inertia” because it signifies the resistance of a substance to a change in temperature Do now… Nelson p20, question set 1.3