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A student heats up a 2.5-g gold block (cs = 0.13 J/g-K) from 13.5 oC to...

A student heats up a 2.5-g gold block (cs = 0.13 J/g-K) from 13.5 oC to 22.8 oC.

a. What is the heat capacity of this gold block?

b. What is the molar heat capacity of gold?

c. How many joules are required to heat this gold block from 13.5 oC to 22.8 oC?

d. If this gold block is submerged into 50 mL of water (cs = 4.18 J/g°-K) at 50 oC after it reaches 22.8 oC, indicate the direction of heat transfer and determine the final temperature of the water and the gold block. (Hint: the final temperature of the water is the same as that of the gold block.)

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Answer #1

(c) Q= mcsT (m- mass of gold block, cs - specific heat capacity, T- Temp. change)

= 2.5 g 0.13 J/g-k 9.3 K = 3.0225 J

(a) Heat capacity C = Q/ T = 3.0225 J/ 9.3K = 0.325 J/K

(b)Molar heat capacity of gold = 0.13 197 = 25.61 J/mol-K

(d) The direction of heat transfer is from water to gold block. initial temp of water= 323K. let the final temp TK. According to calorimetric principle:

mwater C water (323K- T) = m gold C gold (T-295.8)

or, 50 4.18 (323-K) = 2.5 0.13 (T-295.8)

or, 209.325 T = 67603.135 K

or, T = 322.95 K or 49.95 deg centrigade.

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