When 56.0 g of an unknown metal at 30.5 degree C is placed in 83.0 g H_2O at 80.

Question

When 56.0 g of an unknown metal at 30.5 degree C is placed in 83.0 g H_2O at 80.5 degree C, the water temperature decreases to 75.9 degree C. What is the specific heat capacity of the metal? The specific heat capacity of water is 4.184 J/g-K. 25.0 g of ice cubes at 0.0 degree C are combined with 180.0 g of liquid water at 80.0 degree C in a coffee cup calorimeter. Calculate the final temperature reached, assuming no heat loss or gain from the surroundings (Data: specific heat capacity of H_2O(l), c = 4.18 J/g degree C; H_2O(s) rightarrow H_2O(l) delta H = 333 J/g)

Explanation / Answer

1) Heat lost by hot material = heat gained by cold material

=>mm*Cpm*(T-Tm) =mw*Cpw*(Tw-T)

where T =final temperature of the system

Tw =initial temperature of water

Tm =initial temperature of metal

=> Cpm =83*10^-3*4.184*(80.5-75.9) / (56*10^-3*(75.9-30.5)) =0.6283J/kg.K

2)mi*Lfi+mi*Cpw*(T-0) =mw*Cpw*(Tw-T)

=>25*333+ 25*4.18*(T) =80*4.18*(80-T)

=>8325+104.5 *T =30720 -384*T

=>488.5*T =22395 =>T =45.84 oC

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