Question

Liquid A of specific heat capacity 1 J g⁻¹ K⁻¹ at a temperature of ${50}^{\circ }C$ is mixed with 100g of a liquid B of specific heat capacity 2 J g⁻¹ K⁻¹ at ${30}^{\circ }C$.The final temperature of mixture becomes ${40}^{\circ }C$. Find the mass of liquid A.

• A. $333g$
• B. $200g$
• C. $33.3g$
• D. $400g$

Answer 1

The correct option is B

$200g$

Fall in temperature of liquid A = ${10}^{\circ }C$
Rise in temperature of liquid B = ${10}^{\circ }C$
Let m g of liquid A be required.
Heat energy given by m g of liquid A = m$\times$1$\times$(10) J
Heat energy taken by 100g of liquid B = 100 $\times$2$\times$(10) J
Assuming there is no heat loss,
Heat energy given by A = Heat energy taken by B

m$\times$1$\times$(10) = 100 $\times$2$\times$(10)

$m=\frac{100\times 2\times \left(10\right)}{1\times \left(10\right)}$
m= $200g$