Question

A 200g of iron at 120 degrees and a 150 g piece of copper at -50 degrees are dropped into an insulated beaker containing 300 g of ethyl alcohol at 20 degrees. What is the final temperature?

Answer 1

Answer:

T = 15.03°C

Explanation:

given data:

copper specific heat = Sc = 0.385 J/g °C

iron specific iron = Si = 0.450 J/g °C

specific heat of ethanol = Se = 2.46 J/g °C

net heat loss is equal to zero

$(m*S*\Delta T)_{copper} +(m*S*\Delta T)_ {iron} +(m*S*\Delta T)_ {ethanol} = 0$

$150*0.385 *( T - (-50)) + 200*0.450*(T - 120) + 300*2.46 * (T -20) = 0$

$57.75( T - (-50)) + 0.90(T - 120) +738(T -20) = 0$

$57.75T + 2887.5 + 0.90T - 108 + 738T - 14760 = 0$

$57.75T + 0.90T+738T = - 2887.5 + 108+14760$

$796.65T= 11980.5$

T = 15.03°C