Question #203791

1. A chemistry student added 22.5 grams of aluminum at 85.00C to 115 grams of water at 23.00C in a perfect calorimeter. The final temperature of the aluminum-water mixture was 41.40C. Use the student’s data to calculate the specific heat of aluminum in joules/gram0C.

2. Calculate the final temperature that results from mixing 100.0 grams of water at 1000C with 106.0 grams of water at 24.80C

Expert's answer

**Solution.**

"1. m_a=22.5g;"

"t_a=85.0^oC;"

"m_w=115g;"

"t_w=23.0^oC;"

"t=41.4^oC;"

"c_am_a(t_a-t)=c_wm_w(t-t_w);"

"c_a=\\dfrac{c_wm_w(t-t_w)}{m_a(t_a-t)};"

"c_a=\\dfrac{4.2J\/g^oC\\sdot115g\\sdot(41.4^oC-23.0^oC)}{22.5g\\sdot(85.0^oC-41.4^oC)}=9.1J\/g^oC;"

"2. m_1=100.0g;"

"t_1=100^oC;"

"m_2=106.0g;"

"t_2=24.8^oC;"

"c_1m_1(t_1-t)=c_2m_2(t-t_2);"

"m_1t_1-m_1t=m_2t-m_2t_2;"

"t=\\dfrac{m_1t_1+m_2t_2}{m_2+m_1};"

"t=\\dfrac{100.0g\\sdot100^oC+106.0g\\sdot24.8^oC}{106.0g+100.0g}=61.3^oC;"

**Answer: **"1. c_a=9.1J\/g^oC;"

"2. t=61.3^oC."

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