Question

an ice cube at "-10" ⁰c is dropped in a mug containing water at 10 ⁰c.​

Answer 1

Answer:

Here, Mass of water m

w

=100g

Mass of ice, m

i

=10g

Specific heat of water,S

w

=1calg

−1

o

C

−1

Latent heat of fusion of ice,L

fi

=80calg

−1

Let T be the final temperature of the mixture.

Amount of heat lost by water

=m

w

s

w

(△T)

w

=100×1×(50−T)

Amount of heat gained by ice

=m

i

L

fi

+m

i

s

w

(△T)

i

=10×80+10×1×(T−0)

According to principle of calorimetry:

Heat lost = Heat gained

100×1×(50−T)=10×80+10×1×(T−0)

500−10T=80+T

11T=420orT=38.2

o

C

Explanation:

hope it helps you bro have a great day