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Convert the following temperatures: 2. At what temperature will the Celsius and Fahrenheit temperature scales record the same reading? 3. A malaria patient has a body temperature of 39.5°C. Convert th

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4 parts
a

103.1∘F103.1^\circ\text{F}

b

312.65 K312.65 \text{ K}

45.45 m45.45 \text{ m}

1

53.88 J53.88 \text{ J}

2

104.42 K104.42 \text{ K}

Step-by-step solution

12 steps

Here are the solutions to questions 2, 3, 4, and 5.

2. At what temperature will the Celsius and Fahrenheit temperature scales record the same reading?

Step 1: Set the Celsius and Fahrenheit temperature formulas equal to each other.
The formula for converting Celsius to Fahrenheit is TF=95TC+32T_F = \frac{9}{5}T_C + 32.
We want to find the temperature xx where TC=TF=xT_C = T_F = x.

x=95x+32x = \frac{9}{5}x + 32

Step 2: Solve for xx.
Subtract 95x\frac{9}{5}x from both sides:

x−95x=32x - \frac{9}{5}x = 32 55x−95x=32\frac{5}{5}x - \frac{9}{5}x = 32 −45x=32-\frac{4}{5}x = 32

Multiply both sides by −54-\frac{5}{4}:

x=32×(−54)x = 32 \times \left(-\frac{5}{4}\right) x=8×(−5)x = 8 \times (-5) x=−40x = -40

The temperature at which Celsius and Fahrenheit scales record the same reading is −40∘C or −40∘F\boxed{-40^\circ\text{C} \text{ or } -40^\circ\text{F}}.

3. A malaria patient has a body temperature of 39.5∘C39.5^\circ\text{C}. Convert this temperature to (a) ∘F^\circ\text{F} (b) K.

Given temperature TC=39.5∘CT_C = 39.5^\circ\text{C}.

Part a· 2 steps

Convert to Fahrenheit

  1. Use the conversion formula TF=95TC+32T_F = \frac{9}{5}T_C + 32.

    TF=95(39.5)+32T_F = \frac{9}{5}(39.5) + 32

  2. Calculate the Fahrenheit temperature.

    TF=9×7.9+32T_F = 9 \times 7.9 + 32 TF=71.1+32T_F = 71.1 + 32 TF=103.1∘FT_F = 103.1^\circ\text{F}

    The temperature in Fahrenheit is 103.1∘F\boxed{103.1^\circ\text{F}}.

Part b

312.65 K312.65 \text{ K}

Convert to Kelvin

  1. Use the conversion formula TK=TC+273.15T_K = T_C + 273.15.

    TK=39.5+273.15T_K = 39.5 + 273.15

  2. Calculate the Kelvin temperature.

    TK=312.65 KT_K = 312.65 \text{ K}

    The temperature in Kelvin is 312.65 K\boxed{312.65 \text{ K}}.

    4. A steel rod increases its length by 5 mm5 \text{ mm} when the temperature increases by 10∘C10^\circ\text{C}. What is the initial length of the rod if the coefficient of linear expansion for steel is 1.1×10−5 per ∘C1.1 \times 10^{-5} \text{ per } ^\circ\text{C}?

  3. Identify the given values and the formula for linear expansion.

    Change in length, ΔL=5 mm=5×10−3 m\Delta L = 5 \text{ mm} = 5 \times 10^{-3} \text{ m}.
    Change in temperature, ΔT=10∘C\Delta T = 10^\circ\text{C}.
    Coefficient of linear expansion, α=1.1×10−5 per ∘C\alpha = 1.1 \times 10^{-5} \text{ per } ^\circ\text{C}.
    The formula for linear expansion is ΔL=L0αΔT\Delta L = L_0 \alpha \Delta T, where L0L_0 is the initial length.

  4. Rearrange the formula to solve for L0L_0.

    L0=ΔLαΔTL_0 = \frac{\Delta L}{\alpha \Delta T}

  5. Substitute the values and calculate L0L_0.

    L0=5×10−3 m(1.1×10−5 per ∘C)(10∘C)L_0 = \frac{5 \times 10^{-3} \text{ m}}{(1.1 \times 10^{-5} \text{ per } ^\circ\text{C})(10^\circ\text{C})} L0=5×10−31.1×10−4 mL_0 = \frac{5 \times 10^{-3}}{1.1 \times 10^{-4}} \text{ m} L0=51.1×10−1 mL_0 = \frac{5}{1.1 \times 10^{-1}} \text{ m} L0=50.11 mL_0 = \frac{5}{0.11} \text{ m} L0≈45.45 mL_0 \approx 45.45 \text{ m}

    The initial length of the rod is approximately 45.45 m\boxed{45.45 \text{ m}}.

    5. How much heat must be added to a 4.0×10−3 kg4.0 \times 10^{-3} \text{ kg} steel ball bearing in order to increase its temperature by 30 K30 \text{ K} if the specific heat capacity of steel is 4.49×102 J/Kg.K4.49 \times 10^2 \text{ J/Kg.K}? And how much will the temperature of the ball increase if it were made of gold of specific heat capacity 1.29×102 J/Kg.K1.29 \times 10^2 \text{ J/Kg.K} rather than steel?

Part 1

53.88 J53.88 \text{ J}

Heat added to the steel ball

  1. Identify the given values for the steel ball and the formula for heat transfer.

    Mass of steel ball, m=4.0×10−3 kgm = 4.0 \times 10^{-3} \text{ kg}.
    Change in temperature, ΔT=30 K\Delta T = 30 \text{ K}.
    Specific heat capacity of steel, csteel=4.49×102 J/Kg.Kc_{\text{steel}} = 4.49 \times 10^2 \text{ J/Kg.K}.
    The formula for heat transfer is Q=mcΔTQ = mc\Delta T.

  2. Substitute the values and calculate the heat added QQ.

    Q=(4.0×10−3 kg)(4.49×102 J/Kg.K)(30 K)Q = (4.0 \times 10^{-3} \text{ kg})(4.49 \times 10^2 \text{ J/Kg.K})(30 \text{ K}) Q=(4.0×4.49×30)×10−3×102 JQ = (4.0 \times 4.49 \times 30) \times 10^{-3} \times 10^2 \text{ J} Q=(538.8)×10−1 JQ = (538.8) \times 10^{-1} \text{ J} Q=53.88 JQ = 53.88 \text{ J}

    The heat that must be added is 53.88 J\boxed{53.88 \text{ J}}.

Part 2

104.42 K104.42 \text{ K}

Temperature increase for a gold ball with the same heat added

  1. Identify the given values for the gold ball and the heat added.

    Heat added, Q=53.88 JQ = 53.88 \text{ J} (from Part 1).
    Mass of gold ball, m=4.0×10−3 kgm = 4.0 \times 10^{-3} \text{ kg} (assuming the same mass).
    Specific heat capacity of gold, cgold=1.29×102 J/Kg.Kc_{\text{gold}} = 1.29 \times 10^2 \text{ J/Kg.K}.
    The formula for heat transfer is Q=mcgoldΔTgoldQ = mc_{\text{gold}}\Delta T_{\text{gold}}.

  2. Rearrange the formula to solve for ΔTgold\Delta T_{\text{gold}}.

    ΔTgold=Qmcgold\Delta T_{\text{gold}} = \frac{Q}{mc_{\text{gold}}}

  3. Substitute the values and calculate the temperature increase for the gold ball.

    ΔTgold=53.88 J(4.0×10−3 kg)(1.29×102 J/Kg.K)\Delta T_{\text{gold}} = \frac{53.88 \text{ J}}{(4.0 \times 10^{-3} \text{ kg})(1.29 \times 10^2 \text{ J/Kg.K})} ΔTgold=53.88(4.0×1.29)×10−3×102 K\Delta T_{\text{gold}} = \frac{53.88}{(4.0 \times 1.29) \times 10^{-3} \times 10^2} \text{ K} ΔTgold=53.885.16×10−1 K\Delta T_{\text{gold}} = \frac{53.88}{5.16 \times 10^{-1}} \text{ K} ΔTgold=53.880.516 K\Delta T_{\text{gold}} = \frac{53.88}{0.516} \text{ K} ΔTgold≈104.42 K\Delta T_{\text{gold}} \approx 104.42 \text{ K}

    The temperature of the gold ball will increase by approximately 104.42 K\boxed{104.42 \text{ K}}.

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