A 5.80 g sample of SiH4 and an excess of oxygen were placed in a reaction container surrounded by 3.40 kg of water. The initial temperature of the water was 14.10°C. At the end of the reaction, the temperature of the water was 48.90°C. Assume the heat was completely absorbed by the water and no heat was absorbed by the reaction container or the calorimeter or lost to the surroundings. Calculate the heat released in the reaction. The specific heat of water is 4.184 J/g·°C.

q = mass water x specific heat water x (Tfinal-Tinitial)

kilojoules to heat 250kg of gold from 18c to 184c

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To calculate the heat released in the reaction, we can use the formula:

Q = m × c × ∆T

where:
Q is the heat released (in joules),
m is the mass of the substance (in grams),
c is the specific heat capacity of the substance (in J/g·°C), and
∆T is the change in temperature (in °C).

First, let's calculate the mass of water using the given information. We know that the water is 3.40 kg, which is equal to 3400 grams.

Next, we need to calculate the change in temperature (∆T). The initial temperature of the water was 14.10°C, and the final temperature was 48.90°C. So, the change in temperature (∆T) is:

∆T = final temperature - initial temperature
= 48.90°C - 14.10°C
= 34.80°C

Now, we can calculate the heat released (Q) using the formula:

Q = m × c × ∆T
= 3400 g × 4.184 J/g·°C × 34.80°C

Calculating the above expression will give us the value for Q, which represents the heat released in the reaction.