A 129.1 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.

Question

The water increases in temperature to 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 1

Heat Lost = - Heat Gained

General equation q = c * m * change in T

q = heat (J)

c = specific heat capacity (J/ g*K or J/ g*degrees C)

change in T = temperature Tfinal. T initial (K or degrees C)

We know all these pieces of information for the water because we know T initial = 20.7 degrees C, T final = 22.2 degrees C, mass of water = 400 grams and the specific heat capacity of water c = 4.184 J/ g degrees C

the temperature of the water increased so the water is gaining heat

q = c* m * change in T

q = 4.184 * 400 * (22.2-20.7)

q = 4.184 * 400 * 1.5

q gained by water = 2510.4 J

If the water is gaining 2510.4 J then the copper just be losing this same amount of heat

q copper = -2510.4 J

Two objects in thermal equilibrium are at the same temperature so when the temperature of the water stops at 22.2, the temperature of the copper stops as well.

T final = 22.2 degrees C

specific heat capacity copper c = 0.380 J/g degrees C

mass copper = 129.1 grams

This time we have q and our looking for the initial temperature of the copper.

q = c* m * (T final - T initial)

-2510.4 = 0.380 * 129.1 * (22.2 - T initial)

-2510.4 = 49.058 * (22.2 - T initial) divide by 49.058

-51.172 = 22.2 - T initial subtract 22.2

-73.4 = - T initial the negatives cancel our

T initial copper = 73.4 degrees C