A student heated 35.50 grams of metal to 99.0degree C for 10 minutes. The studen
ID: 551408 • Letter: A
Question
A student heated 35.50 grams of metal to 99.0degree C for 10 minutes. The student poured the hot metal into a calorimeter which contained 45.00 grams of water at 25.0 degree C. The temperature of the water rose to 31.0 degree C. Given that the specific heat of water is 4.18 J/g degree C complete the following calculations.
What was the change in temperature of the water?
What was the change in temperature of the metal?
How much heat was gained by the water?
How much heat was lost by the metal?
What is the specific heat of the metal?
Explanation / Answer
Change in teperature of the water = 31-25 = 6 degree clecius
change in temperature of metal = 31- 99 = -68 degree celcius
heat energy gained by water q = m * s * dT
Q = 45 * 4.18 * 6
Q = 1128.6 J
Q system = - Q surroundings
Q water = - Q metal
Q metal = -1128.9 J
Q = m * S * dT
-1128.9 = 35.5 * s *(31-99)
s = specific heat of metal = 0.4676 J/g
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