# If the ring is initially at 99 degrees Celsius, what mass of water at 25 degrees Celsius is needed to lower the ring’s temperature to 38 degrees Celsius?

## A jeweler working with a heated 47g gold ring must lower the ring’s temperature to make it safe to handle. If the ring is initially at 99 degrees Celsius, what mass of water at 25 degrees Celsius is needed to lower the ring’s temperature to 38 degrees Celsius? (ANSWER: 0.0068kg)

Mar 7, 2018

It's 0.0068 kg of water or 6.8 g of water.

#### Explanation:

You'll want to set up a $q = m c \Delta T$ equation (because m'cat needs to cool that ring).
Jokes aside, the change in temperature will be $- {74}^{o} C$, the specific heat capacity for gold is 0.129 (this can be referenced off the internet or found in a table in your textbook), and the mass must be in kilograms .047 kg.

q_"ring"=(m_"ring"c_"ring"DeltaT_"ring")

${q}_{\text{ring}} = .047 \cdot 0.129 \cdot \left(- 61\right) \approx - 0.37 J$

By inserting this in your $q$ equation, a change in energy can be calculated and this loss of energy in the ring (your system) is what the water (your surroundings) will absorb.

${q}_{\text{ring"=-q_"water}}$

You can then substitute the ${q}_{\text{water}}$ for the $m c \Delta T$ of the water. The mass is unknown, the specific heat capacity is 4.184 (again can be referenced from outside source), and the change in temperature is $+ {13}^{o} C$.

q_"ring"=-(m_"water"c_"water"DeltaT_"water")

m_"water"=-(q_"ring")/(c_"water"*DeltaT_"water")

${m}_{\text{water}} = - \frac{- 0.37}{4.184 \cdot 13} \approx 0.0068 k g$ or $6.8 g$