Question

A jeweler working with a heated 47 g gold ring must lower the ring's temperature to make it safe to handle. If the ring is initially at 99ºC, what mass of water at 25ºC is needed to lower the ring's temperature to 38ºC?
5.9 g
3.7 g
2.2 g
6.8 g

Answer 1

Mass of gold m₁ = 47 g

Initial temperature of gold T₁ = 99 C

Specific heat of gold C₁ = 0.129 J/gC

final temperature T₂ = 38 C

Heat needed by the gold to cool down

Q =m₁ * C₁* ( T₁ - T₂)

Q = (47)(0.129)(99-38)

Q = 369.843 J

This heat will be given by the water

we need to find out mass of water m₂

and initial temperature of water is T₃ = 25 C

Specific heat of water C₂ = 4.184 J/gC

Q = m₂*C₂*(T₂ - T₃)

369.843 = m₂(4.184)(38-25)

m₂ = 6.8 g

Answer 2

The answer is D, 6.8 g.