2-kg of ice at -6.0°C is dropped into 5.0 L (5.566-kg) of car coolant. The initial temperature of the coolant is 25.0°C. The ice and coolant are insulated, so energy transfers to the outside is negligible. After hours, all ice melts and the final temperature of the mixture of water-coolant is 1.8°C. The specific heat of ice is 2090 J/kg °C; the specific heat of water is 4186 J/kg °C; the latent heat of fusion for water is 334000 J/kg. A. What is the energy needed for heating the ice to 0 °C? B. Determine the energy needed to melt the ice into water. C. Determine the total energy needed to heat the -6.0°C ice to 1.8°C water. (Please note: the total energy also includes the part heating water from 0.0 °C to 1.8 °C.) D. Determine the specific heat capacity of the coolant.
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.
A.)
To heat 2 kg of ice from -6.0°C to 0°C, we need to use the formula:
Q = m * c * ΔT
where Q is the amount of energy needed, m is the mass of the ice, c is the specific heat capacity of ice, and ΔT is the change in temperature. Substituting the values, we get:
Q = 2 kg * 2090 J/kg °C * (0°C - (-6.0°C))
Q = 26,520 J
Therefore, the energy needed for heating the ice to 0°C is 26,520 J.
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