a) The heat absorbed by the calorimeter and the water can be calculated using the equation q = mcΔT, where q is the heat absorbed, m is the mass of the calorimeter and water, and ΔT is the change in temperature.
q = (81.34 g)(0.385 J/g°C)(36.40°C - 21.00°C) + (200.0 g)(4.184 J/g°C)(36.40°C - 21.00°C)
q = 14,300 J
b) The heat lost to the surroundings can be calculated using the equation q = qin(1 - efficiency), where qin is the heat absorbed by the calorimeter and water and efficiency is the efficiency of the system.
q = 14,300 J(1 - 0.45)
q = 7,841 J
Heat from burning a fuel is absorbed by a copper calorimeter. The calorimeter has a mass of 81.34 g and contains 200.0 g of water initially at 21 degrees Celsius.
a) How much heat was absorbed by the calorimeter and the water if the final temperature of the water is 36.40 degrees Celsius. (c=0.385 for cu, 4.184 J/gc for water)
b) The absorption of heat by the system is 45.00% efficient. How much heat was lost to the surroundings?
I tried it multiple times, the real answers are 14.3 and 7841 kJ and I get 13.3 and 29...Help please?
1 answer