Question

A sample of silver with a mass of 63.3 g is heated to a

temperature of 384.4 K and placed in a container of
water at 290.0 K. The final temperature of the silver
and water is 292.4 K. Assuming no heat loss, what
mass of water was in the container? The specific heat
of water is 4.184 J/(g.°C) and of silver, 0.24 J/(g•°C).

Answer 1

The mass of the water in the container is 139.187g

Step-by-step explanation:

During this process there will not be lost heat. But heat lost of silver is equal to heat won of water

q(lost, silver) = q(gained, water)

-mass * Cpsilver * (T2-T1) = mass * Cpwater * (T2-T1)

⇒mass of silver = 63.3g

⇒Cpsilver = 0.24J/g °C

⇒initial temperature silver = 384.4 K

⇒ final temperature silver = 292.4 K

⇒mass of water = TO BE DETERMINED

⇒Cpwater = 4.184 J/g °C

⇒initial temperature water = 290.0 K

⇒final temperature = 292.4 K

-mass * Cpsilver * (T2-T1) = mass * Cpwater * (T2-T1)

-63.3 * 0.24 * (292.4-384.4) = mass * 4.184 * (292.4 - 290.0)

-1397.664 = mass * 0.24 *2.4

mass water = 139.187g

The mass of the water in the container is 139.187g