A jug of tea is cooled down by putting it in ice. The mass of the tea is 1.8 kg and is initially at 80oC. How many kg of ice, initially at 0oC, are required to bring the mixture to 10oC? (Lf = 3.33 x 105 J/kg, specific heat of water is 4186 J/kg-oC.)
A. 1.2 kg
B. 1.4 kg
C. 1.6 kg
D. 1.8 kg

Question

A jug of tea is cooled down by putting it in ice. The mass of the tea is 1.8 kg and is initially at 80oC. How many kg of ice, initially at 0oC, are required to bring the mixture to 10oC? (Lf = 3.33 x 105 J/kg, specific heat of water is 4186 J/kg-oC.)
	A.	1.2 kg
	B.	1.4 kg
	C.	1.6 kg
	D.	1.8 kg

anonymous · Answer

I think its 1.6

anonymous · Answer

First of all, you should calculate how much calories needed the tea reaches to 10 degrees from 80 degrees. this is easy part, you can use Q=mc(Tf-Ti). This heat energy is given to the ice. Then you should calculate the heat energy that ice takes. Of course, ice melts first then heat up until it gets to 10 degrees.  

i will calculate it for check your answer.

anonymous · Answer

i have got 1.4 kg.. can you show me your math ?