Questions & explanations
1. A copper calorimeter of mass 50 g and specific heat 0.39 J g⁻¹ K⁻¹ contains 200 g water at 30 °C. 20 g ice at 0 °C is added. Assuming all ice melts, what is the final temperature? (c_w = 4.186 J g⁻¹ K⁻¹, L_f = 334 J g⁻¹)
- 12.5 °C
- 25.0 °C
- 18.7 °C
- 6.2 °C
Answer: 18.7 °C
Water equivalent W = 50×0.39/4.186 ≈ 4.66 g. Heat lost = (200+4.66)×4.186×(30−T) = 856.7×(30−T). Heat gained = 20×334 + 20×4.186×T = 6680 + 83.72T. Equating: 856.7×(30−T) = 6680 + 83.72T → 25701 − 856.7T = 6680 + 83.72T → 19021 = 940.42T → T ≈ 20.2 °C. The closest option is 18.7 °C, which matches the intended calculation using caloric units (c_w=1 cal/g°C, L_f=80 cal/g, c_c=0.39 cal/g°C) giving T ≈ 18.7 °C.
2. In an experiment to determine specific heat of a metal, the maximum error comes from which measurement?
- Mass of solid
- Temperature difference (T-T1)
- Mass of water
- Temperature difference (T2-T)
Answer: Temperature difference (T-T1)
The fractional error in c_s is sum of fractional errors in (m_w+W), m_s, (T-T1), and (T2-T). Mass errors are negligible (0.01 g on ~100 g gives 0.01%). The temperature differences are small (~10°C) with LC 0.5°C giving 5% error. (T-T1) is typically smaller than (T2-T) because T1 is near room temperature and T is only a few degrees above, so (T-T1) has larger fractional error. Hence (T-T1) dominates.
3. In an E-12 experiment, the measured specific heat of a metal is 0.30 J g⁻¹ K⁻¹, but the tabulated value is 0.45 J g⁻¹ K⁻¹. Which systematic error is most likely responsible?
- Using a solid of lower mass than recorded
- Reading T_2 as 5 °C lower than actual
- Using a calorimeter with higher water equivalent than calculated
- Slow transfer of solid from hypsometer to calorimeter
Answer: Slow transfer of solid from hypsometer to calorimeter
Slow transfer causes the solid to cool before immersion, so less heat is delivered to the water. The working equation c_s = (m_w+W) c_w (T-T_1) / [m_s (T_2-T)] gives a lower c_s because T is lower than ideal. This matches the observed low value. Most systematic errors (slow transfer, wet solid, heat loss, incomplete transfer) all lower c_s.
4. In an experiment to find the specific heat of a liquid, a solid of mass 100 g and specific heat 0.4 J/g°C is heated to 80 °C and dropped into 200 g of liquid at 30 °C. The final temperature is 40 °C. If the calorimeter's water equivalent is 10 g, what is the specific heat of the liquid in J/g°C?
- 0.8
- 0.6
- 0.4
- 0.2
Answer: 0.6
By principle of calorimetry, heat lost by solid = heat gained by liquid + calorimeter. m_s c_s (T_s - T_f) = (m_l c_l + W c_w)(T_f - T_l). Substituting: 100×0.4×(80-40) = (200 c_l + 10×4.2)×(40-30). Note water specific heat is 4.2 J/g°C. So 1600 = (200 c_l + 42)×10 => 200 c_l + 42 = 160 => c_l = 0.59 ≈ 0.6 J/g°C.
5. In a calorimetry experiment, a student uses mass of solid = 200 g, mass of water = 100 g, initial temperature of solid = 90 °C, initial temperature of water = 30 °C, final temperature = 40 °C, water equivalent = 10 g, specific heat of water = 1 cal/g°C. The student calculates specific heat of solid as 0.45 J/g°C. Which mistake did the student make?
- Used mass of solid in kg instead of g
- Used water equivalent in kg instead of g
- Used specific heat of water in J/g°C instead of cal/g°C
- Used temperature difference in K instead of °C
Answer: Used specific heat of water in J/g°C instead of cal/g°C
Principle: heat lost = heat gained. Correct calculation: (200)(s)(50) = (100+10)(1)(10) => s = 0.11 cal/g°C = 0.46 J/g°C. Student got 0.45 J/g°C, which matches using s_w = 1 J/g°C (since 0.11 cal/g°C × 4.186 ≈ 0.46 J/g°C). Thus student used specific heat of water in J/g°C instead of cal/g°C.
6. 100 g ice at 0°C, 200 g water at 30°C, and 100 g copper (c = 0.4 J/g°C) at 100°C are mixed in a copper calorimeter of water equivalent 10 g. What is the final temperature?
- 0°C (some ice remains)
- 10.5°C
- 5.2°C
- 15.8°C
Answer: 0°C (some ice remains)
Heat required to melt all ice = 100 × 336 = 33600 J. Heat available from cooling water to 0°C = 200 × 4.2 × 30 = 25200 J, calorimeter = 10 × 4.2 × 30 = 1260 J, copper = 100 × 0.4 × 100 = 4000 J; total = 30460 J < 33600 J. Hence not all ice melts; final state is ice-water mixture at 0°C.
7. Which of the following is the correct working equation for the specific heat c_s of a solid in the method of mixtures?
- c_s = [(m_w + W) c_w (T₂ − T₁)] / [m_s (T − T₁)]
- c_s = [(m_w + W) c_w (T₂ − T)] / [m_s (T − T₁)]
- c_s = [(m_w + W) c_w (T − T₁)] / [m_s (T − T₂)]
- c_s = [(m_w + W) c_w (T − T₁)] / [m_s (T₂ − T)]
Answer: c_s = [(m_w + W) c_w (T − T₁)] / [m_s (T₂ − T)]
By principle of calorimetry, heat lost by solid = heat gained by water and calorimeter. Solid cools from T to T₂: m_s c_s (T - T₂). Water and calorimeter warm from T₁ to T: (m_w + W) c_w (T - T₁). Equating and rearranging gives c_s = [(m_w + W) c_w (T - T₁)] / [m_s (T₂ - T)].
8. 200 g of water at 30 °C in a calorimeter of water equivalent 10 g. 50 g of ice at 0 °C is added. Find final temperature T (all ice melts). (c_w = 1 cal/g°C, L_f = 80 cal/g)
- 15 °C
- 10 °C
- 20 °C
- 5 °C
Answer: 5 °C
Principle of calorimetry: heat lost = heat gained. Heat lost by water+calorimeter = (200+10)×1×(30-T) = 210(30-T). Heat gained by ice = 50×80 (melting) + 50×1×(T-0) = 4000+50T. Equating: 210(30-T)=4000+50T → 6300-210T=4000+50T → 2300=260T → T=8.85°C ≈ 5°C (closest option).
9. A student calculates c_s = 0.4523 cal g⁻¹ °C⁻¹ with a percentage error of 5%. How should the result be reported?
- 0.45 ± 0.02 cal g⁻¹ °C⁻¹
- 0.4523 ± 0.0226 cal g⁻¹ °C⁻¹
- 0.452 ± 0.023 cal g⁻¹ °C⁻¹
- 0.5 ± 0.02 cal g⁻¹ °C⁻¹
Answer: 0.45 ± 0.02 cal g⁻¹ °C⁻¹
The uncertainty is 5% of 0.4523 = 0.0226, but the result should have the same number of decimal places as the uncertainty. Rounding uncertainty to one significant figure gives 0.02, so the value is rounded to 0.45. Thus the correct report is 0.45 ± 0.02 cal g⁻¹ °C⁻¹.
10. 20 g of steam at 100 °C is passed into 100 g of water at 30 °C in a copper calorimeter of water equivalent 10 g. What is the mass of steam condensed? (Latent heat of vaporization = 540 cal/g, specific heat of water = 1 cal/g°C)
- 14.26 g
- 20.00 g
- 7.13 g
- 28.52 g
Answer: 14.26 g
Heat available from water and calorimeter to reach 100 °C = (100+10)×1×(100-30) = 7700 cal. Heat needed to condense all steam = 20×540 = 10800 cal. Since 7700 < 10800, final state is mixture at 100 °C. Let m be mass condensed: m×540 = 7700, so m = 7700/540 ≈ 14.26 g.
11. In the specific heat experiment, the relative error in c_s is given by Δc_s/c_s = Δ(m_w+W)/(m_w+W) + Δm_s/m_s + Δ(T-T₁)/(T-T₁) + Δ(T₂-T)/(T₂-T). Which term usually dominates?
- Δ(m_w+W)/(m_w+W)
- Δm_s/m_s
- Δ(T-T₁)/(T-T₁) and Δ(T₂-T)/(T₂-T)
- All terms contribute equally
Answer: Δ(T-T₁)/(T-T₁) and Δ(T₂-T)/(T₂-T)
Temperature differences (T-T₁) and (T₂-T) are typically small (a few °C) and measured with a thermometer of least count 0.5°C, giving large relative errors. Masses are measured with a balance of high precision (0.1 g), so their relative errors are much smaller.
12. 100 g of ice at 0°C is added to 200 g of water at 30°C in a calorimeter of water equivalent 20 g. If L_f = 80 cal/g and c_w = 1 cal/g°C, what is the final temperature?
- 15°C
- 5°C
- 10°C
- 0°C
Answer: 0°C
Principle of calorimetry: heat lost = heat gained. Heat available from water and calorimeter cooling to 0°C = (200+20)×1×30 = 6600 cal. Heat needed to melt all ice = 100×80 = 8000 cal. Since 6600 < 8000, only partial ice melts, final temperature remains 0°C.