1. The hot metal piece weighing 2.500g at 100.0°C is dropped into 40.0 mL water

Question

1. The hot metal piece weighing 2.500g at 100.0°C is dropped into 40.0 mL water at 20.0 °C. The final temperature of the system is 24.0°C.
If the metal is treated as the system, the surrounding is:

a) water

b) styrofoam

c) air

d) metal

2. Is q(system) positive or negative ?

3. Calculate the specific heat of a solid metal (s(metal)), enter numerical value only.

4. According to the definition of q1, q2 and q3, calculate each of these for 5.00 g of ice at -7.77 oC being heated until the temperature of is 15.2 oC
Assuming:
s(ice) = 2.06 J/(goC)
s(water) = 4.184 J/goC
H(fusion) = 333 J/gMelting point of ice = 0 oC (exactly)

q1=?

q2=?

q3=?

Explanation / Answer

1) as the metal is droped into water

metal is system and water is surrounding

2) metal has given heat to water as the temperature of water is increased.

Thus q system is negative.

3) Heat supplied by metal = heat absorbed by water

mass x specific heat of metal x drop in temperature = mass of water x specific heat of water x rise in temperature

Thus 2.5g x s J/K x (100-24) = 40g x 4.18J/K x 4K

thus specific heat of metal s = 3.52J/K

4) H2O(s) ---------------> H2O(s) --------> H2O(l) ------> H2O(l)

5.0 g at -7.77C q1 0 C q2 0C q3 15.2 C

=265.3K 273K 273K 288.2 K

q1 = mass x specific heat of ice x rise in temperature

= 5.0g x2.06 J/g.C. x 7.72 C

= 79.516 J

q2 = mass x deltaHfusion/g

= 5.00g x 333J/g

= 1665 J

q3 =mass x specific heat of water x rise in temperature

= 5.00g x 4.184J/g Cx 15.2C

= 317.99J

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