A 257-g lead ball at a temperature of 78.5° C is placed in a light calorimeter c

Question

A 257-g lead ball at a temperature of 78.5° C is placed in a light calorimeter containing 178 g of water at 24.5° C. Find the equilibrium temperature of the system.

______° C

This is what I did but I got it wrong.. Not sure what I did wrong.

Heat gain = heat loss

( mass of water ) ( Sp heat of water) ( T - T of water ) = (mass of lead ) ( sp heat of lead ) ( T of lead - T )

Where T = equilibrium Temp.

substituting values;

( 178 ) ( 4.184 ) ( T - 24.5 ) = ( 257 ) ( 0.128 ) ( 78.5 - T )

744.75*T -18246.424 = 2582.336-32.896 *T

solving for T

T= 20828.76/711.854

T= 29.25

Explanation / Answer

257 * 0.125(78.5 – T) = 178* 4.18 ( T – 24.5 )

84.2 – T = 23.2T – 567.4

651.6 = 24.2 T

T = 26.9o C

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