Question

A 0.040kg block of copper at 95 degrees celsius is placed in 0.105kg of water at an unknown temperature. After equilibrium is reached, the final temperature is 24 degrees celsius. What was the initial temperature of the water? The specific heat capacity for copper is 0.385J/g degrees celsius

Answer 1

The formula for heat added is:\[Q=cm \Delta T\]where Q is the heat; c is the specific heat capacity of the material and ∆T is the change in temperature.

For the copper we have:\[Q _{emit}=c _{Cu}m _{Cu}\left( T _{2}-T _{1} \right)\]For the water we have:\[Q _{absorb}=c _{H2O}m _{H2O}\left( T2-T1 \right)\]
Now we know that Qemit +Qabsorb=0 since no heat leaves or enters the copper/water system. So we can write:\[c _{Cu}m _{Cu}\left( T _{2}-T _{1}\right)+c _{H2O}m _{H2O}\left( T _{2}-T _{1} \right)=0\]We also know that T2 for the water and the copper are the same since the system arrives at equilibrium. We are given T1 for the copper and need to find T1 for the water. We only need to plug numbers into the last equation and solve it for T1.

Answer 2

You'll likely need to know the following:\[c _{H20}=4.186\frac{ J }{ g°C }\]

Answer 3

Do you see how to find the answer now?