Question

Liquid A of specific heat capacity 0.84 J g⁻¹ K⁻¹ at a temperature of ${50}^{\circ }C$ is mixed with 200 g of a liquid B of specific heat capacity 2.1 J g⁻¹ K⁻¹ at ${30}^{\circ }C$. The final temperature of the mixture becomes ${42}^{\circ }C$. Find the mass of liquid A.

• A. $750g$
• B. $333.33g$
• C. $57.33g$
• D. $400g$

Answer 1

The correct option is A

$750g$

Fall in temperature of liquid A = $50-{42}^{\circ }C$
Rise in temperature of liquid B = $42-{30}^{\circ }C$
Let 'm' g of liquid A be required.
Heat energy lost by ${}^{\prime }{m}^{\prime }g$ of liquid A $=m\times 0.84\times (50-42)J$
Heat energy gained by $200g$ of liquid B $=200\times 2.1\times (42-30)J$
Assuming there is no heat loss,
Heat energy lost by A = Heat energy gained by B
$m\times 0.84\times (50-42)=200\times 2.1\times (42-30)$
$m=\frac{200\times 2.1\times (42-30)}{0.84\times (50-42)}$
$m=750g$