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Social ScienceHow many Cals are required to heat a 28.4 g ice cube from -23.0 °C to -1.0 °C.
How many Cals are required to heat a 28.4 g ice cube from -23.0 °C to -1.0 °C.
Expert Answer
The specific heat capacity for a given substance can be defined as the amount of heat that is needed to raise the temperature of a given substance per unit of the mass. The mathematical relationship between specific heat capacity, mass, change in temperature and the heat of the solution is given by the following equation:
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q=mxcxAT q is the amount of heat released or absorbed during the reaction m is the mass of the given substance c is the specific heat capacity AT is the change in the temperature of the system
In this problem, it has been given that the mass of the ice is given to be 28.4 g.
The change in temperature is given to be from -23oC to -1oC which can be calculated as follows:
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ΔΤ-Τ-Τ --1.0C -(23°C) =22°C
The specific heat capacity of ice is 2.03 J/g oC.
Substituting al...
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q=mxcxAT -28.4 g x2.03 J/g °Cx22°C 1268.344 J
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