Problem 5: Several ice cubes (ρi = 0.9167 g/cm3) of total volume Vi = 215 cm3 and temperature 273.15 K (0.000 °C) are put into a thermos containing Vt = 550 cm3 of tea at a temperature of 313.15 K, completely filling the thermos. The lid is then put on the thermos to close it. Assume that the density and the specific heat of the tea is the same as it is for fresh water (ρw = 1.00 g/cm3, c = 4186 J/kgK). Part (a) Calculate the amount of heat energy Qm in J needed to melt the ice cubes (Lf = 334 kJ/kg). Part (b) Calculate the equilibrium temperature TE in K of the final mixture of tea and water. Part (c) Calculate the magnitude of the total heat transferred QT in J from the tea to the ice cubes.
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.
Problem 5: Several ice cubes (ρi = 0.9167 g/cm3) of total volume Vi = 215 cm3 and temperature 273.15 K (0.000 °C) are put into a thermos containing Vt = 550 cm3 of tea at a temperature of 313.15 K, completely filling the thermos. The lid is then put on the thermos to close it. Assume that the density and the specific heat of the tea is the same as it is for fresh water (ρw = 1.00 g/cm3, c = 4186 J/kgK).
Part (a) Calculate the amount of heat energy Qm in J needed to melt the ice cubes (Lf = 334 kJ/kg).
Part (b) Calculate the equilibrium temperature TE in K of the final mixture of tea and water.
Part (c) Calculate the magnitude of the total
Trending now
This is a popular solution!
Step by step
Solved in 5 steps with 4 images
