Question

pre - laboratory assignment

1. read an authoritative source for a discussion of preparing and interpreting a graph, especially the extrapolation of a curve.
2. a student determined the calorimeter constant of the calorimeter, using the procedure described in this module. the student added 50.00 ml of cold water to 50.00 ml of heated, distilled water in a styrofoam cup. the initial temperature of the cold water was 21.00°c and of the hot water, 29.15°c. the maximum temperature of the mixture was found to be 24.81°c. assume the density of water is 1.00 g ml⁻¹ and the specific heat is 4.184 j g⁻¹ k⁻¹.

(1) determine the δt for the hot water and the cold water.
(2) calculate the heat lost by the hot water.
(3) calculate the heat gained by the cold water.

Explanation:

Step1: Calculate $\Delta T$ for hot water

The formula for $\Delta T$ is $\Delta T=T_{final}-T_{initial}$. For hot water, $T_{initial} = 29.15^{\circ}C$ and $T_{final}=24.81^{\circ}C$.
$$\Delta T_{hot}=24.81 - 29.15=- 4.34^{\circ}C$$

Step2: Calculate $\Delta T$ for cold water

For cold water, $T_{initial} = 21.00^{\circ}C$ and $T_{final}=24.81^{\circ}C$.
$$\Delta T_{cold}=24.81 - 21.00 = 3.81^{\circ}C$$

Step3: Calculate heat lost by hot water

Use the formula $Q = mc\Delta T$. Given $m=
ho V$, $
ho = 1.00g/mL$, $V = 50.00mL$, so $m=1.00\times50.00 = 50.00g$, $c = 4.184Jg^{-1}K^{-1}$, $\Delta T=-4.34^{\circ}C$ (since $1^{\circ}C=1K$).
$$Q_{lost}=50.00\times4.184\times(- 4.34)=-907J$$

Step4: Calculate heat gained by cold water

Using the same formula $Q = mc\Delta T$, $m = 50.00g$, $c = 4.184Jg^{-1}K^{-1}$, $\Delta T = 3.81^{\circ}C$.
$$Q_{gained}=50.00\times4.184\times3.81 = 800J$$

Answer:

(1) Hot water: $-4.34^{\circ}C$, Cold water: $3.81^{\circ}C$
(2) $-907J$
(3) $800J$