Question

A sheet of gold weighing 12.0 g with an initial temperature of 15 °C is placed flat on a 20.0g plate of iron that has been heated to 55°C. What is the final temperature of both pieces of metal? The specific heat of iron is 0.449 J/g°C and the specific heat of gold is 0.129 J/g°C.

Answer 1

49.12 °C

Step-by-step explanation:

`m_(g)` = mass of gold sheet = 12 g

`c_(g)` = specific heat of gold = 0.129 J/g°C

`T_(gi)` = initial temperature of gold sheet = 15 °C

`m_(i)` = mass of iron sheet = 20 g

`c_(i)` = specific heat of iron = 0.449 J/g°C

`T_(ii)` = initial temperature of iron sheet = 55 °C

`T_(f)` = Final equilibrium temperature

Using conservation of heat

`m_(g) c_(g) (T_(f) - T_(gi)) = m_(i) c_(i) (T_(ii) - T_(f) ) \\(12)(0.129)(T_(f) - 15) = (20)(0.449)(55 - T_(f))\\(1.548) (T_(f) - 15) = (8.98) (55 - T_(f))\\T_(f) = 49.12 C`