Question

1. how many joules of heat are needed to raise the temperature of 10.0 g of aluminum from 22°c to 55°c, if the specific heat of aluminum is 0.90 j/g·°c? show your work!!

1. a 15.8 g piece of brass absorbs 1087 joules of heat energy, and its temperature changes from 20°c to 200°c. calculate the specific heat capacity of brass. show your work!!

1. if a 52 g block of an unknown material was originally red - hot (109°c) but is then dropped into a 106 g bucket of cold water (20°c) then you measure the final temperature of the water with the substance in it to be 29°c, what was the specific heat of the material? what material is it? show your work!!

material c
water 4.18
pine wood 2.80
ice 2.09
air 1.00
concrete 0.96
aluminum 0.90
glass 0.84
rock 0.84
iron 0.45
copper 0.39
lead 0.13

heat
m c δt
heat = mcδt

Explanation:

Question 7

Step1: Identify formula and values

We use the heat formula \( Q = mc\Delta T \). Here, \( m = 10.0\space g \), \( c = 0.90\space J/g^\circ C \), initial temp \( T_i = 22^\circ C \), final temp \( T_f = 55^\circ C \). So \( \Delta T = T_f - T_i = 55 - 22 = 33^\circ C \).

Step2: Plug into formula

Substitute values into \( Q = mc\Delta T \): \( Q = 10.0 \times 0.90 \times 33 \).
Calculate: \( 10.0 \times 0.90 = 9.0 \), then \( 9.0 \times 33 = 297 \space J \).

Step1: Recall formula and rearrange

We know \( Q = mc\Delta T \), so solve for \( c \): \( c = \frac{Q}{m\Delta T} \). Given \( Q = 1087\space J \), \( m = 15.8\space g \), \( T_i = 20^\circ C \), \( T_f = 200^\circ C \). \( \Delta T = 200 - 20 = 180^\circ C \).

Step2: Calculate specific heat

Substitute into \( c = \frac{Q}{m\Delta T} \): \( c = \frac{1087}{15.8 \times 180} \).
First, \( 15.8 \times 180 = 2844 \). Then \( c = \frac{1087}{2844} \approx 0.382\space J/g^\circ C \) (approximate value, may vary slightly due to rounding).

Step1: Apply heat transfer principle (heat lost = heat gained)

Let \( c_{unknown} \) be specific heat of unknown. Heat lost by unknown: \( Q_{lost} = m_{unknown}c_{unknown}(T_{unknown, i} - T_f) \). Heat gained by water: \( Q_{gained} = m_{water}c_{water}(T_f - T_{water, i}) \).
Given: \( m_{unknown} = 52\space g \), \( T_{unknown, i} = 109^\circ C \), \( m_{water} = 106\space g \), \( c_{water} = 4.18\space J/g^\circ C \), \( T_{water, i} = 20^\circ C \), \( T_f = 29^\circ C \).
Heat gained by water: \( Q_{gained} = 106 \times 4.18 \times (29 - 20) \).
Calculate \( 29 - 20 = 9^\circ C \). Then \( 106 \times 4.18 = 443.08 \), \( 443.08 \times 9 = 3987.72\space J \).

Step2: Solve for \( c_{unknown} \)

Since \( Q_{lost} = Q_{gained} \), \( 52 \times c_{unknown} \times (109 - 29) = 3987.72 \).
\( 109 - 29 = 80^\circ C \), so \( 52 \times 80 \times c_{unknown} = 3987.72 \).
\( 52 \times 80 = 4160 \). Then \( c_{unknown} = \frac{3987.72}{4160} \approx 0.958\space J/g^\circ C \).
Check the table: Aluminum has \( c = 0.90\space J/g^\circ C \), Concrete has \( 0.96\space J/g^\circ C \), Glass/Rock \( 0.84 \). The value \( \approx 0.96\space J/g^\circ C \) is close to Concrete (0.96) or maybe slight error in calculation. Wait, recalculate:
Wait, \( T_{unknown, i} = 109^\circ C \)? Wait, the problem says "red - hot (109°C)"? Wait, maybe typo? Wait, if it's 100°C? No, as per problem. Wait, let's recalculate:
\( Q_{gained} = m_{water}c_{water}\Delta T_{water} = 106 \times 4.18 \times (29 - 20) \)
\( 29 - 20 = 9 \), \( 106 \times 4.18 = 443.08 \), \( 443.08 \times 9 = 3987.72 \space J \)
\( Q_{lost} = m_{unknown}c_{unknown}(T_{unknown, i} - T_f) = 52 \times c_{unknown} \times (109 - 29) = 52 \times c_{unknown} \times 80 = 4160c_{unknown} \)
Set equal: \( 4160c_{unknown} = 3987.72 \)
\( c_{unknown} = \frac{3987.72}{4160} \approx 0.958 \approx 0.96\space J/g^\circ C \), which matches Concrete (0.96) or close to Aluminum (0.90). Wait, maybe the initial temperature of unknown is 100°C? No, problem says 109°C. Alternatively, maybe I made a mistake. Wait, let's check again:
Wait, mass of bucket? Wait, the problem says "106 g bucket of cold water" – so mass of water is 106 g? Yes. So heat gained by water is correct. Heat lost by unknown: mass 52 g, initial temp 109°C, final temp 29°C, so \( \Delta T = 109 - 29 = 80°C \). So calculation seems right. So specific heat \( \approx 0.96\space J/g^\circ C \), which is close to Concrete (0.96) or maybe a typo in the problem's initial temp. If we assume initial temp of unknown is 100°C, then \( \Delta T = 100 - 29 = 71 \), then \( c = \frac{3987.72}{52\times71} \approx \frac{3987.72}{3692} \approx 1.08\space J/g^\circ C \), which is not in the table. So likely the initial temp is 100°C? No, problem says 109°C. Alternatively, maybe the final temp is 29°C, water mass 106 g, unknown mass 52 g. So with \( c \approx 0.96\space J/g^\circ C \), the material is Concrete (since Concrete has \( c = 0.96\space J/g^\circ C \)).

Step2: Determine material

From the table, Concrete has \( c = 0.96\space J/g^\circ C \), which is close to our calculated \( \approx 0.96\space J/g^\circ C \). So specific heat is approximately \( 0.96\space J/g^\circ C \), and the material is Concrete.

Answer:

297 Joules

Question 8