Answer:
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
