Answer to Question #156033 in General Chemistry for Hannah

Question

I have three questions on my specific heat problems PL worksheet:

1.) What would be the final equilibrium temperature if 80.0g of aluminum at 5.00 degrees Celsius having a specific heat of 0.900J/G degrees celsius is placed in 100.0g of water having an initial temperature of 60.0 degrees celsius?

2.) A bar of lead (specific heat 0.030 J/g-degrees celsius) is heated from 20.4 C to 148.0 C. If this absorbs 324.8 Joules of heat, what is the bar’s mass?

3.) How many joules are released when 80.0 g of water cools from 37.5 degrees celsius to 22.5 degrees celsius? (Specific heat of water is 4.18 J/g*C)

Answer 1

1.

80 × 0.9 × ( 5 -T ) = 100 × 4.18 × ( T - 60 )

= 72 ( 5 -T ) = 418 ( T - 60 ) .

= 360 - 72 T = 418 T - 418 × 60 .

= 418 × 60 - 360 = 418 T - 72 T

= 24720 = 346 T .

= T = 24720 / 346 = 71.44 K .

2. 324.8 = m × 0.03 × ( 148 - 20.4 ) .

m = 324.8 / 0 03 × 127.6 = 84.84 g .

3. Heat = m × S ×delta T .

Heat = 80 × 4.18 × ( 37.5 - 22.5 ) j .

= 5016 j

= 5.016 KJ .