Question

What mass of A liquid a of specific heat capacity 0.84 joule per gram per Kelvin at temperature 40°C must be mixed with 100 g of a liquid B of specific heat capacity 2.1 joule per gram per Kelvin at 20 °C so that the final temperature of mixture become 32 ° C ????​

Answer 1

Explanation:

Let mg of liquid A be required .

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

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

Heat energy given by mg of liquid A

= m × 0.84 × ( 40 - 32 ) J

Heat energy taken by 100 g of liquid B

= 100 × 2.1 × ( 32 - 20 ) J

Assuming that there is no heat loss ,

Heat energy given by A = Heat energy taken by B

m × 0.84 × 8 = 100 × 2.1 × 12

m = 100 × 2.1 × 12 / 0.84 × 8

m = 375 g

Thank you

Answer 2

Explanation:

Mass is equal to 375.

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