a. The quantity of heat in Joules is equal to 1,305.832 Joules.
b. The quantity of heat in Calories is equal to 312.10 Cal.
Given the following data:
- Mass of ice cube = 28.4 grams
- Initial temperature = -23.0°C
- Final temperature = -1.0°C
Scientific data:
- Specific heat capacity of ice = 2.09 J/g°C
To determine the quantity of heat in Joules:
Mathematically, quantity of heat is given by the formula;
[tex]Q=mc\theta[/tex]
Where:
- Q represents the quantity of heat.
- m represents the mass of an object.
- c represents the specific heat capacity.
- ∅ represents the change in temperature.
Substituting the given parameters into the formula, we have;
[tex]Q = 28.4 \times 2.09 \times (-1.0 -[-23.0])\\\\Q=59.356 \times (-1.0+23.0)\\\\Q=59.356 \times22[/tex]
Quantity of heat, Q = 1,305.832 Joules.
In Calories:
[tex]Calories =\frac{1305.832}{4.184}[/tex]
Calories = 312.10 Cal.
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