Calculate the amount of heat a 24 karat gold ring with mass of (1.9000 times 10^
ID: 532000 • Letter: C
Question
Calculate the amount of heat a 24 karat gold ring with mass of (1.9000 times 10^1) g (a very large ring!) would gain when it is heated from 20.0 degree C to (3.89 times 10^1) degree C. [specific heat gold = 0.129 J/g degree C] i.e. -1.0334 times 10 An unknown volume of water at 18.2 degree C is added to (2.680 times 10^1) mL of water at 35.0 degree C. If the final temperature is (2.13 times 10^1) degree C, what was the unknown volume (in mL)? (Assume that no heat was lost to the surroundings; d of water is 1.00 g/mL.) i.e. -1.0334 times 10Explanation / Answer
6) Heat energy needed is calculated by the equation -
q= m c t
q=heat energy gained or lost by a substance
m= mass of gold = 1.9000 x 10^1g
c =specific heat of gold = 0.129 J/°C.
t=change in temperature =(3.89 x 10^1°C -20.0°C)
q = m * c * t
q = 1.9000 x 10^1 * 0.129 J g °C * (3.89 x 10^1°C -20.0°C)
q = 1.9000 x 10^1 * 0.129 J g °C * (1.89 x 10^1°C)
q= 46.3239 J
ANSWER-
4.6324 x 10^1 J
7)
Heat energy needed is calculated by the equation -
q= m c t
Determine energy lost by warm water
q = (2.680 x 10^1 g) (4.186 J/g °C) (2.13 x 10^1 - 35.0 °C )
q = - 1536.9317 J
Determine energy gained by cold water
q = m c t
= m (4.186 J /g °C) (2.13 x 10^1 – 18.20 °C)
= 12.9766 m
Energy lost by warm water = energy gained by cold water
1536.9317 J = 12.9766 m
Mass of cold water = 1536.9317 J / 12.9766
Mass of cold water 118.4387 g
Density of cold water 1.00 g/ml
Volume of cold water ml = Mass of cold water/Density of cold water
118.438 g/1.00g/ml = 118.438ml
ANSwer -
1.1843 x 10^2 ml
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