How long does it take a 740 W coffeepot to bring to a boil 0.68 L of water initi

Question

How long does it take a 740 W coffeepot to bring to a boil 0.68 L of water initially at 8.0°C? Assume that the part of the pot which is heated with the water is made of 369 g of aluminum, and that no water boils away.

min

What will be the equilibrium temperature when a 280 g block of copper at 288°C is placed in a 132 g aluminum calorimeter cup containing 813 g of water at 12.5°C?

°C

A hot iron horseshoe (mass = 0.38 kg), just forged, is dropped into 1.63 L of water in a 0.35 kg iron pot initially at 20°C. If the final equilibrium temperature is 27°C, estimate the initial temperature of the hot horseshoe.

°C

When a 290 g piece of iron at 196°C is placed in a 100 g aluminum calorimeter cup containing 250 g of glycerin at 26°C, the final temperature is observed to be 54°C. What is the specific heat of glycerin?

J/kg·°C

Explanation / Answer

1)

Coffeepot made of aluminum with mass 369 g with a power (P) rating of 740 watts with initial temperature 8°C and

since it is brought to temperature for boiling water its final temperature will be 100°C..

specific heat capacity of Aluminum= 900J/kg K

Water of volume 0.68 litres with initial temperature 8°C and since it is brought to boiling its final temperature will be

100°C

specific heat capacity of Water= 4200J/kg K

The Power is measured in Watts and P=IV

Temperature change = 100°C - 8°C = 92°C


mass x specific heat capacity x temperature change = VIt

P = IV

Therefore,

mass x specific heat capacity x temperature change = Pt
Now,

(0.369 x 900 x 92) + (0.68 x 4200 x 92) = 740 x t

30553.2 + 262752 = 740 x t

t = 396.3584 seconds

t = 6.61 min

2)

mc(tf - ti)water + mc(tf - ti)aluminum= mc(ti - tf)copper

c(water) = 4.186 J/gK

c(aluminum) = 0.897 J/gK

c(copper) = 0.385 J/gK

813(4.186)(tf - 12.5) + 132(0.897)(tf - 12.5) = 280(0.385)(288 - tf)

3403tf - 42540 + 118tf - 1480 = 31046 - 107.8tf

3628.8tf = 75066

tf = 20.69 C

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