A275-g sample of nickel at 100.0°c is placed in 100.0 ml of water at 22.0°c. what is the final temperature of the water? assume that no heat is lost to or gained from the surroundings. specific heat capacity of nickel = 0.444 j/(g·k)

The final temperature of the water is about 39.6°C

Specific Heat Capacity is the amount of energy needed to raise temperature of 1 kg body for 1°C.

Q = Energy ( Joule )

m = Mass ( kg )

c = Specific Heat Capacity ( J / kg°C )

Δt = Change In Temperature ( °C )

Let us now tackle the problem!

Given:

initial temperature of water = t₁ = 22.0°C

specific heat capacity of water = c = 4.186 J/gK

volume of water = V = 100.0 mL

mass of nickel = m₁ = 275 g

initial temperature of nickel = t₁' = 100.0°C

specific heat capacity of nickel = c' = 0.444 J/gK

Asked:

final temperature of the water = t = ?

Solution:

We can calculate the final temperature of the water by using Conservation of Energy as shown below:

Grade: College

Subject: Physics

Chapter: Thermal Physics

Keywords: Heat , Temperature , Block , Aluminium , Ice , Cold , Water

if the thermos has a vacuum innner, then heat through conv currents and conduction are reduced = vacuum flask

okay i have so many questions. but the big one here is: um wt..f is this mess? ?