Question
Asked Oct 15, 2019
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How many Cals are required to heat a 28.4 g ice cube from -23.0 °C to -1.0 °C.

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Expert Answer

Step 1

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:

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
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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

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Step 2

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:

ΔΤ-Τ-Τ
--1.0C -(23°C)
=22°C
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ΔΤ-Τ-Τ --1.0C -(23°C) =22°C

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Step 3

The specific heat capacity of ice is 2.03 J/g oC.

Substituting al...

q=mxcxAT
-28.4 g x2.03 J/g °Cx22°C
1268.344 J
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q=mxcxAT -28.4 g x2.03 J/g °Cx22°C 1268.344 J

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