Question

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

Answer 1

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: $1. c_a=9.1J/g^oC;$

$2. t=61.3^oC.$