recall that heat absorbed or released is given by
Q = mc(T2 - T1)
where
m = mass (in kg)
c = specific heat capacity (in J/kg-K)
T = temperature (in C or K)
*note: Q is (+) when heat is absorbed, and (-) when heat is released.
#1. we just substitute. the c for water is equal to 4.184 J/kg-K (you can google this or look for this in books):
Q = mc(T2-T1)
Q = 3*4.184*(10-80)
Q = -878.64 J
*note that it is negative since it cooled, and thus released heat.
#2,3 and 4: just substitute the given values and solve for the unknown. be careful with units.
hope this helps~ :)
1. How much heat energy is lost by 3 kg of water when it cools from 80 degrees C to 10 degrees C?
2. A 300 g piece of aluminum is heated from 30 degrees C to 150 degrees C. What amount of heat energy is absorbed?
3. Determine the temperature change in each of the following.
(a) 10 kg of water loses 232 kJ of heat energy.
(b) 500 g of copper gains 1.96 kJ of heat energy.
4. After 2 kg of mercury gained 2.52 x 10^4 J of heat energy, its final temperature was 130 degrees C. What was its initail temperature?
5 answers
7 8
Nice but somewhat you would have included the step of all the problems
#1-877,590 J
#1=-877,590 J
1 answer