Respuesta :
Answer: A heat energy of 50160 joules are required to heat 15 grams of ice from [tex]-200^{o}C[/tex] to [tex]1400^{o}C[/tex].
Explanation:
Given : mass of ice = 15 g
Convert grams into kg as follows,
[tex]1 g = 0.001 kg\\15 g = 15 g \times \frac{0.001 kg}{1 g}\\= 0.015 kg[/tex]
Initial temperature = [tex]-200^{o}C[/tex]
Final temperature = [tex]1400^{o}C[/tex]
Formula used to calculate the amount of heat energy is as follows.
[tex]q = m \times C \times (T_{2} - T_{1})[/tex]
where,
q = heat energy
m = mass of substance
C = specific heat of substance (ice here has C = [tex]2090 J/kg ^{o}C[/tex])
[tex]T_{1}[/tex] = initial temperature
[tex]T_{2}[/tex] = final temperature
Substitute the values into above formula as follows.
[tex]q = m \times C \times (T_{2} - T_{1})\\= 0.015 kg \times 2090 J/kg^{o}C \times [1400 - (-200)]^{o}C\\= 50160 J[/tex]
Thus, we can conclude that 50160 joules are required to heat 15 grams of ice from [tex]-200^{o}C[/tex] to [tex]1400^{o}C[/tex].
