A calorimeter of water equivalent 50 g contains 100 g water at 30 ∘ C . In each of the situations, select the option(s) that indicate the correct final temperature.

(A) Warm, melt, then heat. 150 ( 30 − T ) = 15 ( 10 + 80 + T ) T = 11 210 ∘ C (B) Condense, then cool. 150 ( T − 30 ) = 5 ( 540 + 100 − T ) T = 31 1540 ∘ C (C) Mix the water. 150 ( T − 30 ) = 80 ( 70 − T ) T = 23 1010 ∘ C (D) Melt the ice. 150 ( 30 − T ) = 15 ( 80 + T ) T = 20 ∘ C

Start Free Mock LoginAlready registered? Login here

Physics - Heat & Thermo Question with Solution | TestHub

PhysicsHeat & ThermoCalorimetryMedium2 minQB

PhysicsMediummultiple choice

A calorimeter of water equivalent $50 g$ contains $100 g$ water at $30^{\circ} C$ . In each of the situations, select the option(s) that indicate the correct final temperature.

Options:(select one or more)

Answer:

A, B, D

Solution:

(A) Warm, melt, then heat.
$150 (30 - T) = 15 (10 + 80 + T)$
$T = \frac{210}{11}^{\circ} C$
(B) Condense, then cool.
$150 (T - 30) = 5 (540 + 100 - T)$
$T = \frac{1540}{31}^{\circ} C$
(C) Mix the water.
$150 (T - 30) = 80 (70 - T)$
$T = \frac{1010}{23}^{\circ} C$
(D) Melt the ice.
$150 (30 - T) = 15 (80 + T)$
$T = 20^{\circ} C$

Stream:JEESubject:PhysicsTopic:Heat & ThermoSubtopic:Calorimetry

⏱ 2mℹ️ Source: QB

Doubts & Discussion

Loading discussions...