Question

10g of ice at 0° C is mixed with 100g of water at 50°C what is resultant temperature
a) 31.2°C
b) 32.8°C
c) 36.7°C
d) 38.2°C​

Answer 1

Answer:

38.2

Explanation:

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

Answer 2

Answer:

= 100 × 1 × (50 – 0) = 5000 cal. Let temperature of mixture = θ°C.