A thermos flask contains 250 g of coffee at 90° C. To this 20 g of milk at 5° C is added. After equilibrium is established, the temperature of the liquid is (Assume no heat loss to the thermos bottle. Take specific heat of coffee and milk as 1.00 cal / g° C)

Question

A thermos flask contains 250 g of coffee at 90° C. To this 20 g of milk at 5° C is added. After equilibrium is established, the temperature of the liquid is (Assume no heat loss to the thermos bottle. Take specific heat of coffee and milk as 1.00 cal / g° C) (A) 3.23° C (B) 3.15° C (C) 83.7° C (D) 37.8° C

Tardigrade Admin · Accepted Answer

Let after getting equilibrium stage final temperature is T. Then, according to loss of calorimetry Heat loss = Heat gain ⇒ (250)(1)(90-T)=20(1)(T-5) ⇒ T=83.7° C

Q. A thermos flask contains $250 g$ of coffee at $9 0^{\circ} C$ . To this $20 g$ of milk at $5^{\circ} C$ is added. After equilibrium is established, the temperature of the liquid is
(Assume no heat loss to the thermos bottle. Take specific heat of coffee and milk as $1.00 c a l / g^{\circ} C$ )

$3.2 3^{\circ} C$

$3.1 5^{\circ} C$

$83. 7^{\circ} C$

$37. 8^{\circ} C$

Let after getting equilibrium stage final temperature is $T$ .
Then, according to loss of calorimetry
Heat loss $=$ Heat gain
$\Rightarrow (250) (1) (90 - T) = 20 (1) (T - 5)$
$\Rightarrow T = 83. 7^{\circ} C$

Q. A thermos flask contains 250g of coffee at 90∘C. To this 20g of milk at 5∘C is added. After equilibrium is established, the temperature of the liquid is (Assume no heat loss to the thermos bottle. Take specific heat of coffee and milk as 1.00cal/g∘C)

3.23∘C

3.15∘C

83.7∘C

37.8∘C