Question

A hot (70°C) lump of metal has a mass of 250 g and a specific heat of 0.25 cal/g⋅°C. John drops the metal into a 500-g calorimeter containing 75 g of water at 20°C.

The calorimeter is constructed of a material that has a specific heat of 0.10 cal/ g⋅°C.

When equilibrium is reached, what will be the final temperature? cwater = 1.00 cal/g⋅°C.

a. 114°Cb. 72°Cc. 64°Cd. 37°C

Answer 1

d. 37 °C

Step-by-step explanation:

`m_(m)` = mass of lump of metal = 250 g

`c_(m)` = specific heat of lump of metal = 0.25 cal/g°C

`T_(mi)` = Initial temperature of lump of metal = 70 °C

`m_(w)` = mass of water = 75 g

`c_(w)` = specific heat of water = 1 cal/g°C

`T_(wi)` = Initial temperature of water = 20 °C

`m_(c)` = mass of calorimeter = 500 g

`c_(c)` = specific heat of calorimeter = 0.10 cal/g°C

`T_(ci)` = Initial temperature of calorimeter = 20 °C

`T_(f)` = Final equilibrium temperature

Using conservation of heat

Heat lost by lump of metal = heat gained by water + heat gained by calorimeter

`m_(m) c_(m) (T_(mi) - T_(f)) = m_(w) c_(w) (T_(f) - T_(wi)) +  m_(c) c_(c) (T_(f) - T_(ci)) \\(250) (0.25) (70 - T_(f) ) = (75) (1) (T_(f) - 20) + (500) (0.10) (T_(f) - 20)\\T_(f) = 37 C`