Specific Heat (kCal and J) for copper, help please

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To calculate the heat required to raise the temperature of a 50-gram cube of copper from 20°C to 45°C, the specific heat values are essential. The specific heat of copper is 0.092 kcal/kg°C or 385 J/kg°C. The mass of the copper cube is converted to 0.05 kg, and the temperature change (ΔT) is 25°C. Using the formula H = mcΔT, the specific heat must be incorporated correctly to find the total heat in both kcal and Joules. The discussion emphasizes the importance of substituting the known values into the equation for accurate calculations.
strayaway
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Homework Statement


How much heat is required to raise the temperature of a 50 gram cube of copper from 20 degrees Celsius to 45 degrees Celsius? (in kcal and J)
Specific Heat(20degrees Celsius) for Copper (kcal/kg x degrees Celsius) is 0.092. and the Specific Heat for Copper (J/kg x degrees Celsius) is 385.


Homework Equations


H = mc (triangle) T


The Attempt at a Solution


so the first thing I did was convert the 50 grams to 0.05 kg, then I set up the equation by putting "H = 0.05kg x 20degrees x 25 degrees. but then that's where i got stuck because I didn't know how to incorporate the specific heat into this.
 
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strayaway said:

Homework Statement


How much heat is required to raise the temperature of a 50 gram cube of copper from 20 degrees Celsius to 45 degrees Celsius? (in kcal and J)
Specific Heat(20degrees Celsius) for Copper (kcal/kg x degrees Celsius) is 0.092. and the Specific Heat for Copper (J/kg x degrees Celsius) is 385.

Homework Equations


H = mc (triangle) T

The Attempt at a Solution


so the first thing I did was convert the 50 grams to 0.05 kg, then I set up the equation by putting "H = 0.05kg x 20degrees x 25 degrees. but then that's where i got stuck because I didn't know how to incorporate the specific heat into this.

I highlighted a couple of items in your post; the thing you wanted to calculate.

You know m - you even converted it to kg; good.
You know C [you were given its value in 2 unit systems, I would be using the SI unit value to find Joules.
You know delta T [20 to 45 equals a change of ...]

Substitute and calculate.
 
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