Question

You place \(35\;g\) of this cryoprotectant at \(22^\circ C\) in contact with a cold plate that is maintained at the boiling temperature of liquid nitrogen (\(77\;K\)). The cryoprotectant is thermally insulated from everything but the cold plate. Use the values in the table to determine how much heat will be transferred from the cryoprotectant as it reaches thermal equilibrium with the cold plate. (a) \(1.5 \times 1{0^4}\;J\); (b) \(2.9 \times 1{0^4}\;J\); (c) \(3.4 \times 1{0^4}\;J\); (d) \(4.4 \times 1{0^4}\;J\).

Short answer

Therefore, the correct option is (b): \(2.9 \times {10^4}\;{\rm{J}}\)

Step-by-step solution

Write the given data from the question.

Mass of cryoprotectant,\(m = 35\;{\rm{g}}\)

Initial temperature,

\(\begin{array}{l}{T_0} = 22^\circ {\rm{C}}\\{T_0} = 273 + 22\\{T_0} = 295\;{\rm{K}}\end{array}\)

Boiling temperature of the nitrogen,\({T_1} = 77\;{\rm{K}}\)

Determine the formulas to calculate the amount of heat that is transferred from the cryoprotectant as it reaches thermal equilibrium with the cold plate.

The expression to calculate the amount of heat generated cryoprotectant to cool from \(22^\circ {\rm{C}}\) to \( - 20^\circ {\rm{C}}\) is given as follows.

\({Q_1} = m{c_L}\Delta T\) …… (i)

Here,\({c_L}\)is the specific heat of liquid cryoprotectant, and\(\Delta T\)is the change in temperature.

The expression to calculate the amount of heat transferred when cryoprotectant change the phase to change is given as follows.

\({Q_2} = m{L_\gamma }\) …… (ii)

Here,\({L_\gamma }\)is the latent heat of fusion.

The expression to calculate the amount of heat generated cryoprotectant to cool from \( - 20^\circ {\rm{C}}\) to \(77\;{\rm{K}}\;\left( { - 196^\circ {\rm{C}}} \right)\) is given as follows.

\({Q_3} = m{c_S}\Delta T\) …… (iii)

Here,\({c_S}\)is the specific heat of solid cryoprotectant.

The expression to calculate the total heattransferred from the cryoprotectant is given as follows.

\({Q_C} = {Q_1} + {Q_2} + {Q_3}\) …… (iv)

 Step 3: Calculate the amount of heat that is transferred from the cryoprotectant as it reaches thermal equilibrium with the cold plate.

The specific heat of the liquid,\({c_L} = 4.5 \times {10^3}\;{{\rm{J}} \mathord{\left/{\vphantom {{\rm{J}} {{\rm{kg}} \cdot {\rm{K}}}}} \right.\\} {{\rm{kg}} \cdot {\rm{K}}}}\)

The specific heat of the solid,\({c_S} = 2.0 \times {10^3}\;{{\rm{J}} \mathord{\left/{\vphantom {{\rm{J}} {{\rm{kg}} \cdot {\rm{K}}}}} \right.\\} {{\rm{kg}} \cdot {\rm{K}}}}\)

The fusion of water,\({L_\gamma } = 2.8 \times {10^5}\;{{\rm{J}} \mathord{\left/{\vphantom {{\rm{J}} {{\rm{kg}}}}} \right.\\} {{\rm{kg}}}}\)

Calculate the amount of heat generated cryoprotectant to cool from \(22^\circ {\rm{C}}\) to \( - 20^\circ {\rm{C}}\).

Substitute \({\rm{35}}\;{\rm{g}}\) for \(m\),and \(4.5 \times {10^3}\;{{\rm{J}} \mathord{\left/{\vphantom {{\rm{J}} {{\rm{kg}} \cdot {\rm{K}}}}} \right.\\} {{\rm{kg}} \cdot {\rm{K}}}}\) for \({c_L}\)into equation (i).\(\begin{array}{l}{Q_1} = \left( {35 \times {{10}^{ - 3}}\;{\rm{kg}}} \right)\left( {4.5 \times {{10}^3}\;{{\rm{J}} \mathord{\left/{\vphantom {{\rm{J}} {{\rm{kg}} \cdot {\rm{K}}}}} \right.\\} {{\rm{kg}} \cdot {\rm{K}}}}} \right)\left( {295\;{\rm{K}} - 253\;{\rm{K}}} \right)\\{Q_1} = \left( {35 \times {{10}^{ - 3}}\;{\rm{kg}}} \right)\left( {4.5 \times {{10}^3}\;{{\rm{J}} \mathord{\left/{\vphantom {{\rm{J}} {{\rm{kg}} \cdot {\rm{K}}}}} \right.\\} {{\rm{kg}} \cdot {\rm{K}}}}} \right)\left( {42\;{\rm{K}}} \right)\\{Q_1} = 6615\;{\rm{J}}\\{{\rm{Q}}_1} = 6.6\;{\rm{kJ}}\end{array}\)

Calculate the amount of heat transferred when cryoprotectant change the phase to change.

Substitute \(35\;{\rm{g}}\) for \(m\),and \(2.8 \times {10^5}\;{{\rm{J}} \mathord{\left/{\vphantom {{\rm{J}} {{\rm{kg}}}}} \right.\\} {{\rm{kg}}}}\) for \({L_\gamma }\) into equation (ii).

\(\begin{array}{l}{Q_2} = \left( {35 \times {{10}^{ - 3}}\;{\rm{kg}}} \right)\left( {2.8 \times {{10}^5}\;{{\rm{J}} \mathord{\left/{\vphantom {{\rm{J}} {{\rm{kg}}}}} \right.\\} {{\rm{kg}}}}} \right)\\{Q_2} = 98 \times {10^2}\;{\rm{J}}\\{Q_2} = 9.8\;{\rm{kJ}}\end{array}\)

Calculate the amount of heat generated cryoprotectant to cool from \( - 20^\circ {\rm{C}}\) to \(77\;{\rm{K}}\;\left( { - 196^\circ {\rm{C}}} \right)\).

Substitute \(35\;{\rm{g}}\) for \(m\),and \(2.0 \times {10^3}\;{{\rm{J}} \mathord{\left/{\vphantom {{\rm{J}} {{\rm{kg}} \cdot {\rm{K}}}}} \right.\\} {{\rm{kg}} \cdot {\rm{K}}}}\) for \({c_S}\) into equation (iii).

\(\begin{array}{l}{Q_3} = \left( {{\rm{35}} \times {\rm{1}}{{\rm{0}}^{ - 3}}\;{\rm{kg}}} \right)\left( {2.0 \times {{10}^3}\;{{\rm{J}} \mathord{\left/{\vphantom {{\rm{J}} {{\rm{kg}} \cdot {\rm{K}}}}} \right.\\} {{\rm{kg}} \cdot {\rm{K}}}}} \right)\left( {253\;{\rm{K}} - 77\;{\rm{K}}} \right)\\{Q_3} = \left( {{\rm{35}} \times {\rm{1}}{{\rm{0}}^{ - 3}}\;{\rm{kg}}} \right)\left( {2.0 \times {{10}^3}\;{{\rm{J}} \mathord{\left/{\vphantom {{\rm{J}} {{\rm{kg}} \cdot {\rm{K}}}}} \right.\\} {{\rm{kg}} \cdot {\rm{K}}}}} \right)\left( {176\;{\rm{K}}} \right)\\{Q_3} = 12320\;{\rm{J}}\\{{\rm{Q}}_3} = 12.32\;{\rm{kJ}}\end{array}\)

Calculate the total heattransferred from the cryoprotectant.

Substitute \(6.6\;{\rm{kJ}}\) for \({Q_1}\),\({\rm{9}}{\rm{.8}}\;{\rm{kJ}}\) for and \({\rm{12}}{\rm{.32}}\;{\rm{kJ}}\) for \({Q_3}\) into equation (iv). \(\begin{array}{l}Q = 6.6\;{\rm{kJ}} + {\rm{9}}{\rm{.8}}\;{\rm{kJ}} + {\rm{12}}{\rm{.32}}\;{\rm{kJ}}\\Q = \left( {6.6 + {\rm{9}}{\rm{.8}} + {\rm{12}}{\rm{.32}}} \right)\;{\rm{kJ}}\\Q = 2.87 \times {10^4}{\rm{J}}\end{array}\)

Hence the total heattransferred from the cryoprotectant as it reaches thermal equilibrium with the cold plate is \(2.87 \times {10^4}{\rm{J}}\).