Question

what mass of A liquid a of specific heat capacity 0.84 J g^-1 at a temperature 40 °C must be mixed with 100 of a liquid B of specific heat capacity 2.1 J g^-1 k^-1 at 20°C , So that final temperature of mixture become 32°C ?

Answer 1

ANSWERS :-


Let m g of liquid A be required.

Fall in temperature of liquid A = (40 - 32 ) °C

Rise in temperature of liquid B = (32 - 20 ) ° C


Heat energy given by m g of liquid A

= m * 0.84 * ( 40 - 32 ) J .......(I)

Heat energy taken by 100g of liquid B

= 100 * 2.1 * ( 32 - 20 ) J ...... (ii)


Assuming that there is no heat loss,

Heat energy given by A = Heat energy taken by B

or,

m * 0.84 * 8 = 100 * 2.1 *12

or,

m = 100 * 2.1 * 12/0.84 * 8

or,

m = 100 * 21 * 12 *10 / 84 * 8

or,

m = 252000/672

so,

m = 375 g


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❤BE BRAINLY ❤
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