when 20 kilo joule of heat is removed from 1.2 kg of ice originally at -15 degree celsius. its new temperature is
Answers
Answer ⇒ 15.79 °C.
Explanation ⇒ Let the new temperature of the Ice be 't' °C.
Now, Amount of Heat Removed from the Ice = 20 kJ. = 20,000 J.
Mass of the Ice = 1.2 kg.
Specific heat capacity of ice = 2.108 kJ/kgK = 2108 J/kgK.
[Not given in this question, but whenever question will come in exam, it will always be given there.]
Change in temperature = -15 - t
Using the formula,
Q = mcΔt
∴ 2000 = 1.2 × 2108 × (-t - 15)
∴ t + 15 = -0.79
∴ t = -15.79 ° C.
Hence, the final temperature of the ice is -15.79 °C.
Hope it helps.
Answer:Let the new temperature of the Ice be 't' °C.
Now, Amount of Heat Removed from the Ice = 20 kJ. = 20,000 J.
Mass of the Ice = 1.2 kg.
Specific heat capacity of ice = 2.108 kJ/kgK = 2108 J/kgK.
[Not given in this question, but whenever question will come in exam, it will always be given there.]
Change in temperature = -15 - t
Using the formula,
Q = mcΔt
∴ 2000 = 1.2 × 2108 × (-t - 15)
∴ t + 15 = -0.79
∴ t = -15.79 ° C.