Question

Suppose that 28 g of each of the following substances is initially at 27.0 ∘C. What is the final temperature of each substance upon absorbing 2.45 kJ of heat?

a. gold

b. aluminum

c.silver

d.water

Answer 1

specific heat = joule/gm°C
A. Gold = 0.126
B. Silver = 0.233
C. Aluminum = 0.900
D. Water = 4.186

Heat added = specific heat x mass x (t final - t initial)

for gold:
2450J = 0.126 * 28g x ( t final - 27C)
final temp for gold = 721.44 C
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silver:
2450J = 0.233 * 28g x ( t final - 27C)
final temp = 402.53C
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aluminum:
2450J = 0.900 * 28g x ( t final - 27C)
final temp = 124.22 C
---------------------
water:
2450J = 4.186 * 28g x ( t final - 27C)
final temp = 47.90 C