Initially, a beaker has 100 g of water at temperature 90° C . Later another 600 g of water at temperature 20° C was poured into the beaker. The temperature, T of the water after mixing is

Question

Initially, a beaker has 100 g of water at temperature 90° C . Later another 600 g of water at temperature 20° C was poured into the beaker. The temperature, T of the water after mixing is (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Given, Mass of water at 90° C=100 g =100× 10- 3kg Mass of water at 20° C=600 g =600× 10- 3kg From calorimetry, (100 × (10)- 3)Sw(. 90 ° C - T ° C .)=(600 × (10)- 3)Sw(. T ° C - 20 ° C .) ⇒ 90-T=6T-120 T=(210/7)=30° C

Q. Initially, a beaker has 100g of water at temperature 90∘C . Later another 600g of water at temperature 20∘C was poured into the beaker. The temperature, T of the water after mixing is

Given, Mass of water at 90∘C=100g =100×10−3kg Mass of water at 20∘C=600g =600×10−3kg From calorimetry, (100×(10)−3)Sw​(90∘C−T∘C)=(600×(10)−3)Sw​(T∘C−20∘C) ⇒90−T=6T−120 T=7210​=30∘C

Q. Initially, a beaker has $100 g$ of water at temperature $9 0^{\circ} C$ . Later another $600 g$ of water at temperature $2 0^{\circ} C$ was poured into the beaker. The temperature, $T$ of the water after mixing is