Question

A chunk of **gold** weighing **20.0**
grams and originally at **97.53** °C is dropped into
an insulated cup containing **85.0** grams of water at
**21.84**°C.

Assuming that all of the heat is transferred to the water, the
final temperature of the water is °C.

Answer #1

specific heat of gold= 0.126 J/gm.K and that of water= 4.184 J/gm.deg.c

here gold loose heat and water gains heat and this gain and loose contains till equilibrium temperature (t ) is reached.

heat gained by water= mass of water* specific heat of water* change in temperature = 85*4.184*(t- 21.84)=355.64*(t-21.84) (1)

heat lost by gold= 20*0.126*(97.53-t) = 2.52*(97.53-t) (2)

Equatinn Eq.1 and Eq.2

355.64*(t-21.84)= 2.52*(97.53-t)

t*(355.64+2.52)= 2.52*97.53+355.64*21.84

t= 22.37 deg.c

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Silver 0.235 Copper 0.385 Iron 0.449 Aluminum 0.903 Ethanol 2.42
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two parts for one question
-----------------------------------------
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