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Question: The specific heat of aluminum is \[0.214cal{\left( {{g^0}C} \right)^{ - 1}}\] . What is the energy i...

The specific heat of aluminum is 0.214cal(g0C)10.214cal{\left( {{g^0}C} \right)^{ - 1}} . What is the energy in calories necessary to raise the temperature of a 55.5g55.5g piece of aluminium from 23.023.0 to 48.60C{48.6^0}C ?

Explanation

Solution

The energy in calories necessary to raise the temperature of any substance can be calculated from the mass of a substance, the change in temperature and specific heat of a substance. All the values were given and by substituting in the below formula. The energy will be obtained.

Formula used:
q=mCpΔTq = m{C_p}\Delta T
qq is amount of energy in calories
mm is mass of substance
Cp{C_p} is specific heat
ΔT\Delta T is change in temperature

Complete answer:
Given that the specific heat of aluminum (Cp)\left( {{C_p}} \right) is 0.214cal(g0C)10.214cal{\left( {{g^0}C} \right)^{ - 1}}
Specific heat is defined as the amount of energy needed to raise the temperature of one gram of substance through one degree centigrade.
Given substance is aluminium piece with specific heat 0.214cal(g0C)10.214cal{\left( {{g^0}C} \right)^{ - 1}}
The mass of an aluminium piece is 55.5g55.5g
The initial temperature is 23.00C{23.0^0}C and the final temperature is 48.60C{48.6^0}C
Thus, the change in temperature ΔT\Delta T can be obtained by subtracting the initial temperature from final temperature which can be written as 48.623.0=25.60C48.6 - 23.0 = {25.6^0}C
Substitute all the above values in the formula
q=55.5×0.214×25.6=304.05calq = 55.5 \times 0.214 \times 25.6 = 304.05cal
Thus, the amount of energy needed to raise the temperature of a 55.5g55.5g piece of aluminium from 23.023.0 to 48.60C{48.6^0}C is 304.05304.05 calories

Note:
While calculating the amount of energy needed to raise the temperature of a substance the units must be accurate. The specific heat of aluminium is in cal(g0C)1cal{\left( {{g^0}C} \right)^{ - 1}} which means it is written in terms of calories, grams and Celsius. Thus, the mass should be in grams and the temperature must be in Celsius only.