Question

when 20 kilo joule of heat is removed from 1.2 kg of ice originally at -15 degree celsius. its new temperature is​

Answer 1

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 2

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.