In this question, we have to calculate the initial temperature of the glass, and we are provided with the following set of informations:
58.8 grams of glass
0.670 J/g°C specific heat of glass
150.0 grams of water
Initial T of water is 21.0°C
Final T of water is 25.4°C
With these informations, we need to find the total energy that was used to increase the temperature of water from 21 to 25.4 °C. We have to find the energy involved in increasing the temperature of water because this will be the same amount of energy released by the glass sample.
In order to do that, we will use the Calorimetry formula:
Q = mcΔT
Where:
Q = amount of energy
m = mass in grams
c = specific heat capacity, for water is 4.184
ΔT = Final T - Initial T = 25.4 - 21 = 4.4°C
Adding these values into the formula:
Q = 150 * 4.184 * 4.4
Q = 2761 Joules of energy
Now that we have the energy absorbed by water, we can use in the Calorimetry formula, but now to find the initial temperature of glass
Q = mcΔT, we need to turn ΔT into Final T - Initial T
Q = mc(Final T - Initial T)
We have:
Q = -2761, its negative because it is being released, whereas for water is positive because it is being absorbed
m = 58.8 grams
c = 0.670
Final T = 25.4 °C, the final temperature of both water and glass must be the same
Adding these values into the formula:
-2761 = 58.8 * 0.670 (25.4 - Initial T)
-2761 = 39.4 (25.4 - T)
-2761 = 1001 - 39.4T
39.4T = 3762
T = 3762/39.4
T = 95.5 °C is the initial Temperature, rounding to 2 sig fig = 96°C