Question

an unknown mass of each substance, initially at 23.0 °c, absorbs 1.95x10³ j of heat. the final temperature is recorded as indicate. find the mass of each substance.
a. pyrex glass (t_f = 55.4 °c)
b. ethanol (t_f = 44.2 °c)
c. sand (t_f = 62.1 °c)
d. water (t_f = 32.4 °c)

Explanation:

Step1: Identify the formula

The heat - transfer formula is $Q = mc\Delta T$, where $Q$ is the heat absorbed, $m$ is the mass, $c$ is the specific - heat capacity, and $\Delta T$ is the change in temperature.

Step2: Calculate $\Delta T$ for each substance

For Pyrex glass: $\Delta T_{pyrex}=T_f - T_i=55.4 - 23.0 = 32.4^{\circ}C$.
For ethanol: $\Delta T_{ethanol}=T_f - T_i=44.2 - 23.0 = 21.2^{\circ}C$.
For sand: $\Delta T_{sand}=T_f - T_i=62.1 - 23.0 = 39.1^{\circ}C$.
For water: $\Delta T_{water}=T_f - T_i=32.4 - 23.0 = 9.4^{\circ}C$.

Step3: Rearrange the formula to solve for $m$

$m=\frac{Q}{c\Delta T}$.

Step4: Substitute values for each substance

For Pyrex glass: $m_{pyrex}=\frac{1.95\times 10^{3}\ J}{0.75\ J/(g\cdot^{\circ}C)\times32.4^{\circ}C}$.
For ethanol: $m_{ethanol}=\frac{1.95\times 10^{3}\ J}{2.46\ J/(g\cdot^{\circ}C)\times21.2^{\circ}C}$.
For sand: $m_{sand}=\frac{1.95\times 10^{3}\ J}{0.84\ J/(g\cdot^{\circ}C)\times39.1^{\circ}C}$.
For water: $m_{water}=\frac{1.95\times 10^{3}\ J}{4.184\ J/(g\cdot^{\circ}C)\times9.4^{\circ}C}$.

Step5: Calculate the mass

Perform the arithmetic operations to get the mass values for each substance as shown above.

Answer:

a. For Pyrex glass:
The specific - heat capacity of Pyrex glass $c_{pyrex}=0.75 \ J/(g\cdot^{\circ}C)$.
The change in temperature $\Delta T_{pyrex}=T_f - T_i=55.4^{\circ}C - 23.0^{\circ}C = 32.4^{\circ}C$.
Using the heat - transfer formula $Q = mc\Delta T$, we can solve for mass $m$.
$m_{pyrex}=\frac{Q}{c_{pyrex}\Delta T_{pyrex}}=\frac{1.95\times 10^{3}\ J}{0.75\ J/(g\cdot^{\circ}C)\times32.4^{\circ}C}\approx 80.3\ g$

b. For ethanol:
The specific - heat capacity of ethanol $c_{ethanol}=2.46\ J/(g\cdot^{\circ}C)$.
The change in temperature $\Delta T_{ethanol}=T_f - T_i=44.2^{\circ}C - 23.0^{\circ}C = 21.2^{\circ}C$.
Using the heat - transfer formula $Q = mc\Delta T$, we can solve for mass $m$.
$m_{ethanol}=\frac{Q}{c_{ethanol}\Delta T_{ethanol}}=\frac{1.95\times 10^{3}\ J}{2.46\ J/(g\cdot^{\circ}C)\times21.2^{\circ}C}\approx 37.7\ g$

c. For sand:
The specific - heat capacity of sand $c_{sand}=0.84\ J/(g\cdot^{\circ}C)$.
The change in temperature $\Delta T_{sand}=T_f - T_i=62.1^{\circ}C - 23.0^{\circ}C = 39.1^{\circ}C$.
Using the heat - transfer formula $Q = mc\Delta T$, we can solve for mass $m$.
$m_{sand}=\frac{Q}{c_{sand}\Delta T_{sand}}=\frac{1.95\times 10^{3}\ J}{0.84\ J/(g\cdot^{\circ}C)\times39.1^{\circ}C}\approx 59.9\ g$

d. For water:
The specific - heat capacity of water $c_{water}=4.184\ J/(g\cdot^{\circ}C)$.
The change in temperature $\Delta T_{water}=T_f - T_i=32.4^{\circ}C - 23.0^{\circ}C = 9.4^{\circ}C$.
Using the heat - transfer formula $Q = mc\Delta T$, we can solve for mass $m$.
$m_{water}=\frac{Q}{c_{water}\Delta T_{water}}=\frac{1.95\times 10^{3}\ J}{4.184\ J/(g\cdot^{\circ}C)\times9.4^{\circ}C}\approx 49.7\ g$