Question

1. syrup has a specific heat capacity of 1.3 j/g.°c. if 200 g of syrup absorbs 5000 j of heat energy to reach a final temperature of 55°c, what was its initial temperature?
2. what is the mass of water if it absorbs 65000 j of heat energy to go from a temperature of 75°c to 94°c?
3. what is honeys specific heat if 90.0 g are heated for 18.0 minutes and experience a temperature change of 40.0°c absorbing 7500.0 j of heat?
4. water’s specific heat is 4.19 j/g.°c. olive oil’s specific heat is 1.9 j/g.°c.

a. which one would get hotter faster?
b. which one would keep its heat for longer?
c. why would we put water in a car’s cooling system over olive oil?

1. when preparing tea, naomi pours 205 g of boiling water into a porcelain cup. the cup also weighs 205 g. naomi wants to find the specific heat of the porcelain cup.

she obtained the following results:

• before pouring the water into the tea cup: - after pouring the water into the cup and stirring gently:
• temperature of cup = 25°c - final temperature of water = 88°c
• initial temperature of water = 105°c

calculate the specific heat capacity of the cup.

Explanation:

Step1: Recall the heat - transfer formula

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

Step2: Solve for the initial temperature of the syrup (Question 6)

We know that $Q = 5000\ J$, $m = 200\ g$, $c=1.3\ J/g^{\circ}C$, and $T_f = 55^{\circ}C$.
First, we can re - arrange the formula $Q = mc\Delta T$ to solve for $\Delta T$: $\Delta T=\frac{Q}{mc}$.
Substitute the values: $\Delta T=\frac{5000\ J}{200\ g\times1.3\ J/g^{\circ}C}=\frac{5000}{260}\ ^{\circ}C\approx19.23^{\circ}C$.
Since $\Delta T=T_f - T_i$, then $T_i=T_f-\Delta T$.
$T_i = 55^{\circ}C-19.23^{\circ}C = 35.77^{\circ}C$.

Step3: Solve for the mass of water (Question 7)

We know that $Q = 65000\ J$, $c = 4.18\ J/g^{\circ}C$, $T_i = 75^{\circ}C$, and $T_f = 94^{\circ}C$.
First, find $\Delta T=T_f - T_i=94^{\circ}C - 75^{\circ}C = 19^{\circ}C$.
From $Q = mc\Delta T$, we can solve for $m$: $m=\frac{Q}{c\Delta T}$.
Substitute the values: $m=\frac{65000\ J}{4.18\ J/g^{\circ}C\times19^{\circ}C}=\frac{65000}{79.42}\ g\approx818.44\ g$.

Step4: Solve for the specific heat of honey (Question 8)

We know that $Q = 7500\ J$, $m = 90.0\ g$, and $\Delta T = 40.0^{\circ}C$.
From $Q = mc\Delta T$, we can solve for $c$: $c=\frac{Q}{m\Delta T}$.
Substitute the values: $c=\frac{7500\ J}{90.0\ g\times40.0^{\circ}C}=\frac{7500}{3600}\ J/g^{\circ}C = 2.083\ J/g^{\circ}C$.

Step5: Answer questions about specific heat comparison (Question 9)

a. The substance with the lower specific heat capacity will get hotter faster. Since the specific heat of olive oil ($1.9\ J/g^{\circ}C$) is lower than that of water ($4.19\ J/g^{\circ}C$), olive oil will get hotter faster.
b. The substance with the higher specific heat capacity will keep heat for longer. Water has a higher specific heat capacity, so water will retain heat for longer.
c. We put water in a car's cooling system because it can absorb a large amount of heat without a large increase in temperature due to its high specific heat capacity.

Step6: Solve for the specific heat of the porcelain cup (Question 10)

The heat lost by water is equal to the heat gained by the cup.
Heat lost by water: $Q_{water}=m_{water}c_{water}(T_{i,water}-T_f)$.
Here, $m_{water}=205\ g$, $c_{water}=4.18\ J/g^{\circ}C$, $T_{i,water}=105^{\circ}C$, and $T_f = 88^{\circ}C$.
$Q_{water}=205\times4.18\times(105 - 88)=205\times4.18\times17=205\times71.06 = 14567.3\ J$.
Heat gained by the cup: $Q_{cup}=m_{cup}c_{cup}(T_f - T_{i,cup})$.
Here, $m_{cup}=205\ g$, $T_{i,cup}=25^{\circ}C$, $T_f = 88^{\circ}C$.
Since $Q_{water}=Q_{cup}$, we have $14567.3\ J=205\ g\times c_{cup}\times(88 - 25)$.
$14567.3\ J=205\ g\times c_{cup}\times63$.
$c_{cup}=\frac{14567.3}{205\times63}=\frac{14567.3}{12915}\approx1.13\ J/g^{\circ}C$.

Answer:

Question 6: $35.77^{\circ}C$
Question 7: $818.44\ g$
Question 8: $2.083\ J/g^{\circ}C$
Question 9a: Olive oil
Question 9b: Water
Question 9c: Because water has a high specific - heat capacity and can absorb a large amount of heat without a large temperature increase.
Question 10: $1.13\ J/g^{\circ}C$