<p>A <span style=\"color: #dd0000;\">198</span>-g aluminum calorimeter contains

Question

<p>A <span>198</span>-g aluminum calorimeter contains <span>490</span> g of water at <span>19.0</span><span>&#176;</span> C. Aluminum shot with a mass equal to <span>273</span> g is heated to 100.0&#176; C and is then placed in the calorimeter. Find the final temperature of the system, assuming that there is no heat transfer to the surroundings.</p>

Explanation / Answer

Q.

A 198g aluminum calorimeter contains 490 g of water at 19.0 C. Aluminum shot with a mass equal to 273 g is heated to 100.0 C and is then placed in the calorimeter. Find the final temperature of the system, assuming that there is no heat transfer to the surroundings?

A.

heat lost by Al shot = heat gained by calorimeter

=> 273*C_Al * (100-t) = 198*(C_water) *(t-19)  +  490*(C_Al)*(t-19)

The specific heat of aluminum is 0.214 cal/g·°C.

The specific heat of water is 1 cal/g·°C.

solve for t 

=> 273*0.214(100-t) = 198*0.214(t-19) + 490*1(t-19)

=> t = 27 degrees

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