Answer to Question #101106 in Chemistry for Jeffery

Question

a) A student places 289 g of water in a calorimeter at 14.9° C. She then drops in a piece of silver at 87.5 degrees C. If the final temperature of the water is 19.7°C, what is the mass of the silver? (cAg = 0.0558 cal/g°C)
b) Later, her lab partner pours 200.6 g of water at 95.8°C into a 300.9 g silver bowl at 18.4°C. What will the final temperature be (assuming that no heat is transferred to or from the environment)?(algebra first)
c) Realizing that her prior experiment was imperfect, the first student decides to find the heat capacity of the calorimeter. She places 165 g of water at 18.5°C in a calorimeter. She then adds 85 g of water at 84.7°C. If the final temperature of the system is 39.5°C, what is the heat capacity of the calorimeter?
d) Her partner then places 285.5 g of water at 91.6°C and puts it in the calorimeter. He then drops in 16.5 g of ice at -3.5°C.What will the final temperature of the system be? ( HΔ fus of ice = 79.7 cal/g, c of ice = 0.50 cal/g°C)

Answer 1

a) Initial conditions : $m_{w}=289g;T_{w_i}=14.9° C;T_{Ag_i}=87.5° C$

Final conditions : $T_{w_f}=T_{Ag_f}=19.7° C$

Heat lost by system = Heat gained by system

$\implies m_{Ag}s_{Ag}\Delta T_{Ag}= m_{w}s_{w}\Delta T_{w}$

$\implies m_{Ag}=[289*1*(19.7-14.9)]/$$[0.0558*(87.5-19.7)]$

$=(289*4.8)/(0.0588*67.8)$

$=347.962g.$

b) Initial conditions: $m_{w}=200.6g;m_{Ag}=300.9g;$$T_{w_i}=95.8° C;T_{Ag_i}=18.4° C$

Let final equilibrium temperature be $T$

Heat lost by system = Heat gained by system

$\implies m_{w}s_{w}\Delta T_{w}=m_{Ag}s_{Ag}\Delta T_{Ag}$

$\implies 200.6*1*( 95.8-T)=$$300.9*0.0558*(T-18.4)$

$\implies 19217.48-200.6T=16.79T-308.94$

$\implies T=18908.539/217.39$

$=86.98^{\omicron}C$

c) $T_{c_i}=18.5°C;T_{c_f}=39.5°C$

Let heat capacity of calorimeter be C $cal/^{\omicron}C$

Heat lost by system = Heat gained by system

$\implies 85*1*(84.7-39.5)=$$[165*1*(39.5-18.5)]+C*(39.5-18.5)$

$\implies C*21=(85*45.2)-(165*21)$

$\implies C=377/21$

$=17.952 cal/^{\omicron}C$

d) Let final equilibrium temperature be $T (T>0^{\omicron}C)$

Heat lost by system = Heat gained by system

$285.5*1*(91.6-T)+ C_{calorimeter}(91.6-T)=$$16.5*0.5*(0-(-3.5))+$$m*\Delta H_{fusion_{ice}}+16.5*1*(T-0)$

$\implies (285.5+17.952)*(91.6-T)=$$(16.5*0.5*3.5)+(16.5*79.7)+16.5*T$

$\implies 27796.238-303.45T=16.5T+1343.925$

$\implies 319.95T=26452.313$

$\implies T= 82.676^{\omicron}C$