Question

For each trial, enter the amount of heat lost by the metal, qmetal.

Hint: The specific heat of water is 4.184 J/goC. Be careful of your algebraic sign here and remember that the change in temperature is equal to the final temperature minus the initial temperature. Your answer should be reported to 4 digits.
Note: You should always carry 1 or 2 extra digits beyond the number of significant figures until your final calculation.
Trial # MASS Metal Ti metal Mass cool water Ti water Tr q metal
#1: 183.428 101.2 70.001 24.2 38.2
#2: 183.430 101.0 70.008 24.0 37.8
#3 183.430 100.7 70.271 23.2 36.6

Answer 1

Trial 1: Heat lost by the metal = -4100 J

Trial 2: Heat lost by the metal = -4042 J

Trial 3: Heat lost by the metal = -3940 J

Step-by-step explanation:

Assuming that all of the heat lost by the metal is transferred to the water and no heat is lost to the surroundings; Heat lost by metal = heat gained by water. Qmetal = -Qwater

Note: If the experiment was carried out in a calorimeter of known heat capacity, some heat is lost to the calorimeter.

Hence, Qmetal = -(Qwater + Qcalorimeter)

Quantity of heat, q = mass * specific heat capacity * temperature change

Trial 1

mass of water = 70.001 g

specific heat capacity of water = 4.184 J/g/°C

Initial temperature of water = 24.2 °C

final temperature of mixture = 38.2 °C

Temperature change of water = 38.2 - 24.2 = 14.0 °C

Heat gained by water = -(70.001 * 4.184 * 14.0) = -4100 J

Therefore, heat lost by the metal = -4100 J

Trial 2

mass of water = 70.008 g

specific heat capacity of water = 4.184 J/g/°C

Initial temperature of water = 24.0 °C

final temperature of mixture = 37.8 °C

Temperature change of water = 37.8 - 24.0 = 13.8 °C

Heat gained by water = -(70.008 * 4.184 * 13.8) = -4042 J

Therefore, heat lost by the metal = -4042 J

Trial 3

mass of water = 70.271 g

specific heat capacity of water = 4.184 J/g/°C

Initial temperature of water = 23.2 °C

final temperature of mixture = 36.6 °C

Temperature change of water = 36.6 - 23.2 = 13.4 °C

Heat gained by water = -(70.271 * 4.184 * 13.4) = -3940 J

Therefore, heat lost by the metal = -3940 J