Question

the calorimeter contains 900. g of water and the temperature change is 8.32 °c. the heat capacity of the dry calorimeter is 2240 j/°c. the specific heat of water is 4.18 j/g °c. what is the heat transfer of the calorimeter? $q_{cal} = q_{h_2o} + q_{dry}$ $q_{cal} = ? j$ enter either a + or - sign and the magnitude.

Explanation:

Step1: Calculate \( q_{H_2O} \)

The formula for heat transfer of water is \( q = mc\Delta T \), where \( m = 900. \, \text{g} \), \( c = 4.18 \, \text{J/g}^\circ\text{C} \), \( \Delta T = 8.32^\circ\text{C} \).
\( q_{H_2O} = 900. \times 4.18 \times 8.32 \)
\( q_{H_2O} = 900. \times 34.8 \) (approx, after \( 4.18\times8.32\approx34.8 \))
\( q_{H_2O} = 31320 \, \text{J} \) (exact: \( 900\times4.18\times8.32 = 900\times34.8016 = 31321.44 \, \text{J} \))

Step2: Calculate \( q_{dry} \)

The formula for heat transfer of dry calorimeter is \( q = C\Delta T \), where \( C = 2240 \, \text{J/}^\circ\text{C} \), \( \Delta T = 8.32^\circ\text{C} \).
\( q_{dry} = 2240 \times 8.32 \)
\( q_{dry} = 18636.8 \, \text{J} \)

Step3: Calculate \( q_{cal} \)

\( q_{cal} = q_{H_2O} + q_{dry} \)
\( q_{cal} = 31321.44 + 18636.8 \)
\( q_{cal} = 49958.24 \, \text{J} \approx 5.00\times10^4 \, \text{J} \) (or keep more decimals). Since heat is absorbed (assuming temperature increase, so + sign), but let's check the calculation precisely.

Wait, more precise calculation:
\( q_{H_2O} = 900 \times 4.18 \times 8.32 \)
\( 4.18\times8.32 = 4.18\times(8 + 0.32) = 33.44 + 1.3376 = 34.7776 \)
\( 900\times34.7776 = 31300.84 \) (wait, 900*34.7776: 34.7776*900 = 34.7776*9*100 = 313.0084*100 = 31300.84? Wait no, 900*34.7776: 34.7776*900 = 34.7776*9*100 = (313.0084)*100 = 31300.84? Wait, 4.18*8.32: 4*8.32=33.28, 0.18*8.32=1.4976, total 33.28+1.4976=34.7776. Then 900*34.7776= 34.7776*900: 34*900=30600, 0.7776*900=700.8, total 30600+700.8=31300.8? Wait, earlier I miscalculated 4.18*8.32. Let's do 4.18*8.32:

4.18 * 8.32:
= (4 + 0.18) * (8 + 0.32)
= 4*8 + 4*0.32 + 0.18*8 + 0.18*0.32
= 32 + 1.28 + 1.44 + 0.0576
= 32 + 1.28=33.28; 33.28+1.44=34.72; 34.72+0.0576=34.7776. Correct. Then 900*34.7776= 34.7776*9*100= 313.0084*100= 31300.84 J.

\( q_{dry}=2240*8.32=2240*8 + 2240*0.32=17920 + 716.8=18636.8 J \)

Then \( q_{cal}=31300.84 + 18636.8=49937.64 J \approx 4.99\times10^4 J \), or 49938 J. If we consider significant figures: mass is 900. (3 sig figs), specific heat 4.18 (3), delta T 8.32 (3), C 2240 (maybe 4 sig figs? 2240 could be 3 or 4; 900. is 4? Wait 900. has a decimal, so 4 sig figs? Wait 900. g is 4 sig figs (the decimal indicates that the trailing zero is significant). So 900. (4), 4.18 (3), 8.32 (3), 2240 (let's assume 4 sig figs, as 2240 could be written as 2.240×10³, so 4). Then:

\( q_{H_2O}=900. \times 4.18 \times 8.32 = 900. \times 34.7776 = 31300.84 \, \text{J} \) (4 sig figs in 900., 3 in others, so result should have 3 sig figs? Wait, 900. is 4, 4.18 is 3, 8.32 is 3. When multiplying, the result has the least number of sig figs, which is 3. So 31300.84 ≈ 31300 J (3 sig figs? Wait 31300 with 3 sig figs is 3.13×10⁴. But maybe the problem doesn't care about sig figs, just calculation.

Wait, the problem says "Enter either a + or - sign AND the magnitude." Assuming the temperature increased, so heat is absorbed, so +. So the total heat transfer is \( q_{cal} = +49938 \, \text{J} \) (or more precisely, as calculated).

Wait, let's recalculate:

\( q_{H_2O} = m \times c \times \Delta T = 900 \, \text{g} \times 4.18 \, \text{J/g}^\circ\text{C} \times 8.32^\circ\text{C} \)
\( = 900 \times 4.18 \times 8.32 \)
\( 4.18 \times 8.32 = 34.7776 \)
\( 900 \times 34.7776 = 31300.84 \, \text{J} \)

\( q_{dry} = C \times \Delta T = 2240 \, \text{J/}^\circ\text{C} \times 8.32^\circ\text{C} = 18636.8 \, \text{J} \)

\( q_{cal} = 31300.84 + 18636.8 = 49937.64 \, \text{J} \approx 49900 \, \text{J} \) (or 5.00×10⁴ J if we round to three significant figures, since 4.18 and 8.32 have three, 900. has four, 2240 has four; the least is three, so 5.00×10⁴ J, or 4.99×10⁴ J).

But let's check the exact sum: 31300.84 + 18636.8 = 49937.64, which is approximately 49900 J or 5.00×10⁴ J. The sign is positive because the calorimeter is absorbing heat (assuming the reaction is exothermic, so calorimeter gains heat, hence +q).

Answer:

\( +49900 \, \text{J} \) (or more precisely \( +49938 \, \text{J} \), or \( +5.00\times10^4 \, \text{J} \))