You are thinking ahead to spring when one of your friends is having an outdoor wedding. Your plan is to design the perfect lemonade for the event. The problem with lemonade is that you make it in room temperature and then add ice to cool it to a pleasant 10 oC. Usually, the ice melts diluting the lemonade too much. To help you solve this problem, you look up the specific heat capacity of water (1.0 cal/g oC), the specific heat capacity of ice (0.5 cal/g oC), and the latent heat of fusion of water (80 cal/g). You assume that the specific heat capacity of lemonade is the same as water. Since you will cool your lemonade in a Thermos jug, assume no heat is added to the lemonade from the environment. Using that information, you calculate how much water you get from all the ice melting if you make 6 quarts (5.6 kg) of lemonade at room temperature (23 oC) and add ice which comes straight from the freezer at -5.0 oC.
Energy transfer
The flow of energy from one region to another region is referred to as energy transfer. Since energy is quantitative; it must be transferred to a body or a material to work or to heat the system.
Molar Specific Heat
Heat capacity is the amount of heat energy absorbed or released by a chemical substance per the change in temperature of that substance. The change in heat is also called enthalpy. The SI unit of heat capacity is Joules per Kelvin, which is (J K-1)
Thermal Properties of Matter
Thermal energy is described as one of the form of heat energy which flows from one body of higher temperature to the other with the lower temperature when these two bodies are placed in contact to each other. Heat is described as the form of energy which is transferred between the two systems or in between the systems and their surrounding by the virtue of difference in temperature. Calorimetry is that branch of science which helps in measuring the changes which are taking place in the heat energy of a given body.
You are thinking ahead to spring when one of your friends is having an outdoor wedding. Your plan is to design the perfect lemonade for the event. The problem with lemonade is that you make it in room temperature and then add ice to cool it to a pleasant 10 oC. Usually, the ice melts diluting the lemonade too much. To help you solve this problem, you look up the specific heat capacity of water (1.0 cal/g oC), the specific heat capacity of ice (0.5 cal/g oC), and the latent heat of fusion of water (80 cal/g). You assume that the specific heat capacity of lemonade is the same as water. Since you will cool your lemonade in a Thermos jug, assume no heat is added to the lemonade from the environment. Using that information, you calculate how much water you get from all the ice melting if you make 6 quarts (5.6 kg) of lemonade at room temperature (23 oC) and add ice which comes straight from the freezer at -5.0 oC.
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