 Juan Hewlett

2022-01-06

0.250 kilogram of water at ${75.0}^{\circ }C$ are contained in a tiny, inert beaker. How much ice, at a temperature of $-{20.0}^{\circ }C$ degrees, needs to be added to the water to reach the desired temperature of ${40.0}^{\circ }C$ degrees? Jacob Homer

The final temperature of the system is $40{C}^{\circ }$ and the total heat flow for the system will be Zero.
${Q}_{\to tal}={Q}_{ice}+{Q}_{water}=0$ (1)
The ice will go through three stages, so you need to be careful while you are calculating ${Q}_{ice}$, the first stage includes the change of the temperature of the ice but without any change in the phase ''''it still solid through this stage”, the second stage starts when the ice reaches its melting point and ends when the all the ice melts, the final stage includes the changing of the temperature of the water that was ice before melting, meaning that this water has the same mass as the ice. Hence, ${Q}_{ice}$ can be written as follows
${Q}_{ice}={m}_{ice}×{c}_{ice}×\mathrm{△}{T}_{ice}+{m}_{ice}×{L}_{f}+{m}_{ice}×{c}_{water}×\mathrm{△}{T}_{me
Substitute for the quantities in the above equation to get ${Q}_{ice}$
${Q}_{ice}={m}_{ice}\left[\left(2100\frac{J}{kg}\cdot C°\right)×\left[0-\left(-20C°\right)\right]+\left(33.4×{10}^{4}\frac{J}{kg}\right)+\left(4190\frac{J}{kg}\cdot C°\right)×\left(40C°-0\right)\right]$
${Q}_{ice}=\left[5.436×{10}^{5}\frac{J}{kg}\right]{m}_{ice}$
Now, let's calculate Q for the water which has a mass of (0.25 kg) and its temperature drops from 75C° to 40C°
${Q}_{water}={m}_{water}×{c}_{water}×\mathrm{△}T$
${Q}_{water}=\left(0.25kg\right)×\left(4190\frac{J}{kg}\cdot C°\right)×\left(40C°-75C°\right)$
${Q}_{water}=-3.67×{10}^{4}J$
Now, to get the value of mice, substitute for ${Q}_{water}$ and ${Q}_{ice}$ into eq.(1)
$\left[5.436×{10}^{5}J\right]{m}_{ice}-3.67×{10}^{4}\frac{J}{kg}=0$
${m}_{ice}=\frac{3.67×{10}^{4}J}{5.436x{10}^{5}\frac{J}{kg}}$
${m}_{ice}=0.0675kg$ Maricela Alarcon

Required
the mass of the
$Ice\left({m}_{ice}\right)$
Given
${M}_{water}=0.250k{g}_{ww}{T}_{1water}=75{C}_{\left\{ww\right\}}^{\circ }{T}_{1ice}=-20C{°}_{ww}{T}_{2}=40C°$
Solution
Final temperature be $40{C}^{\circ }$ mean that the ice contain to Heat quantity as following
$-20{C}^{\circ }\stackrel{{Q}_{ice}}{⇒}0C°$
$0C°\stackrel{{Q}_{ice-water}}{⇒}40C°$
So the thermal heat of the system ${Q}_{t}$, will be
${Q}_{t}={Q}_{ice}+{Q}_{ice-water}+{Q}_{water}+{Q}_{water}=0$
Where ${Q}_{water}$ is for the water cooled from
and ${Q}_{water}$ is the water heated from
${m}_{ice}×{C}_{ice}×\delta {T}_{ice}+{m}_{ice}×{L}_{f}+{m}_{w}×C\left\{w\right\}×\delta {T}_{w}+{m}_{w}×C\left\{w\right\}×\delta {T}_{w}=0$ (1)
by solving equation (1) ${m}_{ice}$
${m}_{ice}=\frac{{m}_{w}×C\left\{w\right\}×\delta {T}_{w}+{m}_{w}×C\left\{w\right\}×\delta {T}_{w}}{{C}_{ice}×\delta {T}_{ice}+{L}_{f}}$ karton

Explanation:
Heat gained by ice in taking the total temperature to ${40}^{\circ }C=$ Heat lost by the water
Total Heat gained by $ice=Heat$ used by ice to move from
Heat used to melt at ${0}^{\circ }C+$ Heat used to reach ${40}^{\circ }C$ from ${0}^{\circ }C$
To do this, we require the specific heat capacity of ice, latent heat of ice and the specific heat capacity of water. All will be obtained from literature.
Specific heat capacity of $ice={C}_{i}=2108{\frac{J}{kg.}}^{\circ }C$
Latent heat of $ice=L=334000\frac{J}{kg}$
Specific heat capacity of $water=C=4186\frac{J}{kg.}°C$
Heat gained by ice in taking the total temperature to ${40}^{\circ }C=m{C}_{i}\mathrm{△}T+mL+mC\mathrm{△}T=m\left(2108\right)\left(0-\left(-20\right)\right)+m\left(334000\right)+m\left(4186\right)\left(40-0\right)=42160m+334000m+167440m=543600m$
Heat lost by $water=mC\mathrm{△}T=0.25\left(4186\right)\left(75-40\right)=36627.5J$
543600m=36627.5
m=0.0674kg=67.4g of ice.

Do you have a similar question?