Question

A 2.61 g lead weight, initially at 11.1 ∘C, is submerged in 7.67 g of water at 52.6 ∘C in an insulated container. What is the final temperature of both the weight and the water at thermal equilibrium?

Answer 1

Answer:

Equilibrium temperature will be $T=52.2684^{\circ}C$

Explanation:

We have given weight of the lead m = 2.61 gram

Let the final temperature is T

Specific heat of the lead c = 0.128

Initial temperature of the lead = 11°C

So heat gain by the lead = 2.61×0.128×(T-11°C)

Mass of the water m = 7.67 gram

Specific heat = 4.184

Temperature of the water = 52.6°C

So heat lost by water = 7.67×4.184×(T-52.6)

We know that heat lost = heat gained

So $2.61\times 0.128\times (T-11)=7.67\times 4.184\times (52.6-T)$

$0.334T-3.67=1688-32.031T$

$T=52.2684^{\circ}C$