Question

a 99.1 g piece of copper (specific heat 0.380 j/g·°c) is heated and then placed into 400.0 g of water initially at 20.7°c. the water increases in temperature to a final temperature of 22.2°c.
what is the initial temperature (in °c) of the copper? (the specific heat of water is 4.184 j/g·°c).
answer:
°c

Explanation:

Step1: Calculate heat gained by water

The formula for heat \(q = mc\Delta T\). For water, \(m = 400.0\space g\), \(c = 4.184\space J/g\cdot^{\circ}C\), \(\Delta T=T_{final}-T_{initial}=22.2 - 20.7=1.5^{\circ}C\)
\(q_{water}=400.0\times4.184\times1.5\)
\(q_{water}=2510.4\space J\)

Step2: Relate heat lost by copper to heat gained by water

By the principle of heat exchange \(q_{copper}=-q_{water}\). For copper, \(m = 99.1\space g\), \(c = 0.380\space J/g\cdot^{\circ}C\), \(T_{final}=22.2^{\circ}C\), let \(T_{initial}\) of copper be \(x\)
\(q_{copper}=99.1\times0.380\times(22.2 - x)\)
Since \(q_{copper}=- 2510.4\)
\(99.1\times0.380\times(22.2 - x)=-2510.4\)

Step3: Solve for \(x\)

First, calculate \(99.1\times0.380 = 37.658\)
The equation becomes \(37.658\times(22.2 - x)=-2510.4\)
\(22.2 - x=\frac{-2510.4}{37.658}\)
\(22.2 - x=- 66.66\)
\(x=22.2 + 66.66\)
\(x = 88.9^{\circ}C\)

Answer:

\(88.9^{\circ}C\)