A sample of water with a mass of 27.56g and an unknown temperature loses 2443 Joules. If the final temperature is found to be 62.50 C. What was the initial temperature?

41.3 °C

Explanation:

From the question given above, the following data were obtained:

Mass (M) of water = 27.56 g

Heat (Q) loss = 2443 J

Final temperature (T2) = 62.5 °C

Initial temperature (T1) =?

NOTE: The specific heat capacity (C) of water is 4.18 J/g°C

Thus, we can obtain the initial temperature of the water by using the following formula:

Q = MC(T2 – T1)

2443 = 27.56 × 4.18 (62.5 – T1)

2443 = 115.2008 (62.5 – T1)

Divide both side by 115.2008

2443 / 115.2008 = (62.5 – T1)

21.20645 = 62.5 – T1

Collect like terms

21.20645 – 62.5 = – T1

– 41.3 = – T1

Divide both side by – 1

– 41.3 /– 1= – T1 / –1

41.3 = T1

T1 = 41.3 °C

Thus, the initial temperature of the water was 41.3 °C

the maximum mass of nh3 is nh3 in grams