q = m x C x T Where: q = heat absorbed or released m = mass C= specific heat of
ID: 984939 • Letter: Q
Question
q = m x C x T
Where:
q = heat absorbed or released
m = mass
C= specific heat of the substance in joules/g °C or in calories/g °C
T = change in temperature
The chemist often works with the unit of energy known as the joule, but it is common in the US to refer to calories rather than joules when looking at energy. In the problems below, we will determine q, the heat absorbed or released, in calories rather than in joules. To this end, the specific heats in the problems in this section will be given in the units of calories/g °C instead of joules/g °C.
A 1.4-gram cashew nut is burned. The heat released by the cashew is completely transferred to 89.5 grams of water and raises the temperature of the water from 21.0°C to 64.0°C. a) Use the equation above to determine the heat absorbed by the water in calories! (Cwater= 1.00 calorie/g °C) (If you feel you need assistance with this problem, check out example 14.1 in your text.) Click to add text
b) Convert that answer to kilocalories (kcal). Remember that the dietary charts are in capital ‘C’ calories or kcal. So the number in kcal is the number of Calories that the water absorbed. Click to add text
c) According to the problem, the energy that raised the temperature of the water came from the cashew that completely burned. If the 1.4 gram cashew nut released that many Calories or kcal, how many Calories are released per gram of cashew?
Explanation / Answer
a) q = m x C x T
Given that m = 1.4 gram
C = specific heat of water = 1.00 calorie/g °C
T = rise in temperature = 64.0°C - 21.0°C = 43 °C
q = m x C x T
= 1.4 g x 1.00 calorie/g °C x 43 °C
= 60.2 cal
Therefore, heat absorbed by the water = 60.2 cal
b) 60.2 cal = 0.0602 kcal
c) From part a,
1.4 gram cashew nut released the 60.2 calories.
Calories are released per gram of cashew = 60.2 calories / 1.4 = 43 calories
Therefore,
43 clories are released per gram of cashew.
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